Power Series PDF

# Power Series ## Definition The series of the form: $Σ_{n=0}^∞ a_n(x-x_0)^n = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + ... + a_n(x-x_0)^n$ is called a power series. Where, 'x' is a continuous variable and the constants $a_n$ are real and independent of x. ## Properties In this section we shall derive...

# Power Series ## Definition The series of the form: $Σ_{n=0}^∞ a_n(x-x_0)^n = a_0 + a_1(x-x_0) + a_2(x-x_0)^2 + ... + a_n(x-x_0)^n$ is called a power series. Where, 'x' is a continuous variable and the constants $a_n$ are real and independent of x. ## Properties In this section we shall derive the properties of functions which are represented by power series. - $f(x) = Σ_{n=0}^∞ C_n x^n$ - More generally: $f(x) = Σ_{n=0}^∞ C_n(x-a)^n$ These are called **analytic functions**. If the series converges for all $x$ in $(-R, R)$, for some $R_0 > 0$ (R may be +∞), we say that 'f' is expanded in a power series about the point $x=0$. Similarly, If the series converges for $| x-a | < R$, 'f' is said to be expanded in a power series about the point $x=a$. As a matter of convenience we shall often take $a=0$ without any loss of generality. ## Radius of convergence The number 'R' is called the radius of convergence of the power series $Σ_{n=0}^∞ a_nx^n$, and the set of all $x$ for which $| x | < R$ (i.e., open interval $(-R, R)$ is called the interval of convergence. ## Examples - $Σ_{n=0}^∞ a_nx^n$ converges for $| x | < 1 $ & diverges for $| x | > 1$ - $Σ_{n=0}^∞ x^n$ converges for $-1 ≤ x <1$ & diverges elsewhere. - $Σ_{n=0}^∞ \frac{x^n}{n}$ converges absolutely for $| x | ≤ 1$ & diverges for $| x | > 1$ **Therefore:** $$ R=lim_{n→∞} sup| \frac{a_n}{a_{n+1}} | $$ **Example** Prove that the power series: $$ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $$ has infinite radius of convergence. **Solution:** Here, the given series ($Σ_{n=0}^∞ a_nx^n$) is: $a_n = \frac{1}{n!}$ The radius of convergence is given by the formula: $$ R=lim_{n→∞} \frac{a_n}{a_{n+1}} $$ Therefore: $$ R=lim_{n→∞} \frac{1}{n!} \frac{(n+1)!}{1} $$ $$ R=lim_{n→∞} (n+1) $$ $$ R=∞ $$ Thus, the given power series converges absolutely for all x. ## Theorem If a power series $Σ_{n=0}^∞ a_nx^n$ converges for $|x| < R$ and let us define a function f(x) = $Σ_{n=0}^∞ a_nx^n$ for $|x| < R$ then $Σ_{n=0}^∞ a_nx^n$ converges uniformly on $[-R+ε, R-ε]$ no matter which $ε > 0$ is choosen and that the function f is continous and differentiable on $(-R,R)$ and $f'(x) = Σ_{n=1}^∞ na_nx^{n-1}$ for $|x| ≤ R$. **Proof:** Let $ε > 0 $ be any no. given for $|x| < R-ε$, we have: $|a_nx^n| ≤ |a_n|(R-ε)^n$ But since $Σ_{n=0}^∞ |a_n|(R-ε)^n$ coverges absolutely. (every power series converges absolutely within its interval of convergence) Therefore, by W.M. test the series $Σ_{n=0}^∞ a_nx^n$ converges uniformly on $[-R+ε, R-ε]$. Again, since every term of the series $Σ_{n=0}^∞ a_nx^n$ is continous and differentiable on $(-R, R)$ and $Σ_{n=0}^∞ a_nx^n$ is uniformly covergent on $[-R+ε, R-ε]$ therefore, its sum function f is also continous & differentiable on $(-R, R)$. Also, $lim_{n→∞} |na_n|^{\frac{1}{n}} = lim_{n→∞} (n)^{\frac{1}{n}} | a_n|^{\frac{1}{n}}=\frac{1}{R}$ Hence, the differentiated series $Σ_{n=1}^∞ na_nx^{n-1}$ is also a power series and has the same radius of convergence R as $Σ_{n=0}^∞ a_nx^n$. Therefore, $Σ_{n=1}^∞ na_nx^{n-1}$ is uniformly convergent on $[-R+ε, R-ε]$. Hence, $f'(x) = Σ_{n=1}^∞ na_nx^{n-1}$, $|x| < R$. ## Note Since each term of the power series is continuous and integrable on $(-R,R)$ and the power series $Σ_{n=0}^∞ a_nx^n$ is uniformly convergent on $[-R+ε, R-ε]$, therefore, the sum function f is continuous and integrable on the interval. Moreover, $lim_{n→∞} | \frac{a_n}{a_{n+1}} |^{\frac{1}{n}}=\frac{1}{R}$ So that, integrated series also has the same radius of convergence as $Σ_{n=0}^∞ a_nx^n$. Therefore, the above theorem can include the case that $Σ_{n=0}^∞ a_nx^n$ is integrable & the integrable series which is also a power series has the same radius of convergence. Moreover, by the repeated application of the theorem, f can be differentiated oe integrated any no. of times. ## Corollary Under the hypothesis of the above theorem, f has derivative of all orders in (-R,R) which are given by, $f^(k)(x) = Σ_{n=k}^∞ n(n-1)...(n-k+1) C_nx^{n-k}$ In particular, $f^(k)(0) = K!C_k$, ($C_k = 0, 1, 2, ...$) (Here, f^(o) means f & $f^(k)$ is the $K^{th}$ derivative of f, for K = 1, 2, 3, ...) **Proof:** $f(x) = a_0 + a_1x +a_2x^2 + ... + a_nx^n +...$ By the above theorem, f(x) is differentiable any no. of times. Let us differentiate K times. $f^(1)(x) = a_1 + 2a_2x + 3a_3x^2 + ... + na_nx^{n-1} +...$ $f^(2)(x) = 2.1.a_2 + 3.2.a_3x + 4.3.a_4x^2 + ... +n(n-1)a_nx^{n-2} +...$ ... $f^(3)(x) = 3!a_3 + 4.3.2.a_4x + ... + n(n-1)(n-2).a_nx^{n-3} +...$ $f^(4)(x) = 4!a_4 + 5.4.3.a_5x + ... + n(n-1)(n-2).(n-3)a_nx^{n-4} +...$ ... $f^(k)(x) = k!a_k + (k+1)k(k-1)2a_{k+1}x + ... + n(n-1)...(n-k+1)a_nx^{n-k} +...$ $f^(k)(x) = Σ_{n=k}^∞ n(n-1)...(n-k+1)a_nx^{n-k}$ Also, $f^(k)(0) = K!a_k$ the other terms vanishing at $x=0$ **If the series $Σ_{n=0}^∞ a_nC_n$ converges at an endpoint say at $x=R$, then f is continuous on only in (-R,R) but also at $x= R$. This follows from Abel's theorem (for simplicity of notation, we take $R=1$).** ## **Abel's theorem**: Suppose Σcn converges (Put: $$f(x) = Σ_{n=0}^∞ c_nx^n (1-x<1)$$ Then $lim_{x→1}f(x) = Σ_{n=0}^∞ c_n$ **Proof**: Let $S_n=c_0+c_1+...+c_n$, $S_{-1} = 0$ Then: $$f(x) = (1-x)Σ_{n=0}^∞ S_n x^n$$ For $| x | <1 $, we let $m→∞$ and $lim_{m→∞}S_m = S$, and: $$Σ_{n=0}^∞ S_n x^n$$ is 2nd term zeros and: $$(1-x^m)=1$$ Suppose $lim_{m→∞}S_m=S$. Let ε>0 be given. Choose N so that $|S_n-S| ≤ ε$ for $n ≥ N$ Then since $(1-x^{m+1})=1$ for $|x| < 1$ $$|f(x)-S| = |(1-x)Σ_{n=0}^{m-1}(S_n-S)x^n|$$ $$≤ (1-x)Σ_{n=0}^{N-1}|S_n-S|x_n + (1-x)Σ_{n=N}^m|S_n-S|x^n $$ But: $$Σ_{n=0}^{N-1}|S_n-S|x^n ≤ Σ_{n=0}^{N-1}|S_n-S|$$ And: $$(1-x)Σ_{n=N}^m|S_n-S| ≤ ε · Σ_{n=N}^m x^n ≤ ε$$ by (2) Since (1-x)Σ_{n=N}^m|S_n-S| is a convergent function for having zero value at $n=1$ $:|f(x)- S| ≤ ε (Σ_{n=0}^{N-1}|S_n-S| + ε $$ $lim_{x→1-}|f(x) - S| ≤ ε (Σ_{n=0}^{N-1}|S_n-S| + ε $$ $lim_{x→1-}|f(x) - S| ≤ ε (Σ_{n=0}^{N-1}|S_n-S| + ε $$ ## Examples on Power Series 1. **Prove that the power series: $1 + x + x^2 + x^3 + ... $ has infinite radius of convergence.** **Solution:** Here the given series ($Σ_{n=0}^∞ a_nx^n$) is: $$a_n = 1$$ The radius of convergence is given by the formula: $$R = lim_{n→∞} \frac {a_n}{a_{n+1}}$$ Therefore: $$ R=lim_{n→∞} \frac{1}{1} $$ $$ R=lim_{n→∞} 1 $$ $$ R=1 $$ Thus, the given series converges for any value of x other than 0. 2. **Prove that the power series $1 + 2x + 2!x^2 + 3!x^3 +....$ fails to converges for any value of x other than 0.** **Solution:** Here the given series is: $$ Σ_{n=0}^∞ a_nx^n = 1 + x + 2!x^2 + 3!x^3 +.... $$ $$ a_n = n! $$ The radius of convergence is given by: $$ R = lim_{n→∞} \frac{a_n}{a_{n+1}} $$ Therefore: $$ R = lim_{n→∞} \frac{n!}{(n+1)!} $$ $$ R = lim_{n→∞} \frac {1}{(n+1)} $$ $$ R = 0 $$ Thus, the given series converges for any value of x other than 0. 3. **Find the interval of absolute convergence for the power series: $2 + \frac{x}{2^2} + \frac{x^2}{2^3} + ... $** **Solution:** Here the given power series is: $$ Σ_{n=1}^∞ a_nx^n = \frac{x}{2^2} + \frac{x^2}{2^3} + ... $$ The radius of convergence R is: $$ \frac{1}{R} = lim_{n→∞} sup|a_n|^{\frac{1}{n}} $$ Therefore: $$ \frac{1}{R} = lim_{n→∞} sup| \frac{1}{2^{n+1}} |^{\frac{1}{n}} $$ $$ \frac{1}{R} = lim_{n→∞} \frac{1}{2} $$ $$ R = 2. $$ Thus, the given power series converges absolutely for all $|x| < 2.$ 4. **Find the radius of convergence of the following series: i) $Σ_{n=0}^∞ \frac{(2n)!}{(n!)^2} x^n$ ii) $Σ_{n=1}^∞ \frac{n^n}{(n!)^2} x^n$ iii) $Σ_{n=0}^∞ \frac{x^n}{n^2}$ iv) $Σ_{n=0}^∞ n^2 x^n$** **Solution** i) Here the given power series is: $Σ_{n=0}^∞ \frac{(2n)!}{(n!)^2} x^n$ Comparing with $Σ_{n=0}^∞ c_n x^n$, we have $c_n = \frac{(2n)!}{(n!)^2}$ By ratio test, we have: $$ R = \frac{1}{lim_{n→∞} sup | \frac{c_{n+1}}{c_n} |} $$ Therefore: $$ R = \frac{1}{lim_{n→∞} sup | \frac{(2(n+1))!}{((n+1)!)^2} \frac{(n!)^2}{(2n)!} |} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac{(2n+2)(2n+1)}{(n+1)(n+1)} |} = \frac{1}{lim_{n→∞} | \frac{(2n+2)(2n+1)}{(n+1)(n+1)} |} $$ $$ R = \frac{1}{lim_{n→∞} | \frac {4n^2 + 6n + 2} {n^2 + 2n + 1} |} = \frac{1}{4} $$ Therefore, the radius of converge is $R = \frac{1}{4}$. ii) Here the given power series is: $Σ_{n=1}^∞ \frac{n^n}{(n!)^2} x^n$ Comparing with $Σ_{n=0}^∞ c_n x^n$, we have $c_n = \frac{n^n}{(n!)^2}$ Therefore: $$ R = \frac{1}{lim_{n→∞} sup| \frac{c_{n+1}}{c_n} |} $$ $$ R=\frac{1}{lim_{n→∞} sup | \frac{(n+1)^{n+1}}{((n+1)!)^2} \frac{(n!)^2}{n^n} |} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac{(n+1)(n+1)^n}{(n+1)^2 (n!)^2} \frac{(n!)^2}{n^n} | } $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac{(n+1)^n}{n^n (n+1)} |} = \frac{1}{lim_{n→∞} | \frac{(1+\frac{1}{n})^n}{(n+1)} | } $$ $$ R = \frac{1}{lim_{n→∞} | \frac{e}{(n+1)} |} = e $$ Therefore, the radius of converge is R = e. iii) Here the given power series is: $Σ_{n=0}^∞ \frac{x^n}{n^2}$ Comparing with $Σ_{n=0}^∞ c_n x^n$, we have $c_n = \frac {1} {n^2}$. Therefore: $$ R = \frac{1}{lim_{n→∞} sup| \frac{c_{n+1}}{c_n} |} $$ $$ R = \frac{1}{lim_{n→∞} sup| \frac{1}{(n+1)^2} \frac{n^2}{1} |} $$ $$ R = \frac{1}{lim_{n→∞} | \frac {n^2}{(n+1)^2} |} = 1 $$ Therefore, the radius of converge is R = 1. iv) Here, the given power series is: $Σ_{n=0}^∞ n^2 x^n$ on comparing with $Σ_{n=0}^∞ c_n x^n$, we have $c_n = n^2$ Therefore: $$ R = \frac{1}{lim_{n→∞} sup | \frac{c_{n+1}}{c_n} |} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac { (n+1)^2 } {n^2} |} = \frac{1}{lim_{n→∞} | \frac {n^2 + 2n + 1} {n^2} |} $$ $$ R = \frac{1}{lim_{n→∞} | \frac {1 + \frac{2}{n} + \frac{1}{n^2} } {1} |} = 1 $$ Therefore, the radius of converge is R = 1. 5. **Show that the radius of convergence of the power series $Σ_{n=0}^∞ \frac{z^n}{(n+1)^2(n+2)^2} (x+1)^n$ is 1** **Soln:** Here the given series is: $Σ_{n=0}^∞ \frac{1}{(n+1)^2(n+2)^2} (x+1)^n$ on comparing with $Σ_{n=0}^∞ c_n x^n$, we have $c_n = \frac{1}{(n+1)^2(n+2)^2}$ Therefore: $$ R = \frac{1}{lim_{n→∞} sup |\frac{c_{n+1}}{c_n}|} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac{1}{(n+2)^2(n+3)^2} \frac{(n+1)^2(n+2)^2}{1} |} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac { (n+1)^2 (n+2)^2} {(n+2)^2 (n+3)^2}|} = \frac{1}{lim_{n→∞} |\frac{(n+1)^2}{(n+3)^2}|} $$ $$ R = \frac{1}{lim_{n→∞} | \frac{n^2 + 2n + 1}{n^2 + 6n + 9} |} = \frac {1}{lim_{n→∞} | \frac {1 + \frac {2}{n} + \frac{1}{n^2}}{1 +\frac {6}{n} + \frac{9}{n^2}} |} $$ $$ R = 1 $$ Hence, proved. ## Example: True or False. **Prove that the following power series is convergent for $ -1 < x ≤ 1$: $$ \frac{2}{1} - \frac{3}{3} + \frac{5}{5} - \frac {7}{7} + ... $$** **Solution:** Here, given series is: $$ Σ_{n=1}^∞ \frac{(-1)^{n-1}}{2n-1} x^{2n-1} $$ On comparing with $Σ_{n=1}^∞ a_n x^n$, we have $a_n = \frac{(-1)^{n-1}}{2n-1}$ The radius of convergence is given by: $$ R = \frac{1}{lim_{n→∞} sup | \frac{a_{n+1}}{a_n} |} $$ Therefore: $$ R = \frac{1}{lim_{n→∞} sup | \frac{(-1)^n}{2(n+1)-1} \frac{2n-1}{(-1)^{n-1}} |} $$ $$ R = \frac{1}{lim_{n→∞} sup | \frac{2n-1}{2n+1} |} $$ $$ R = \frac{1}{lim_{n→∞} | \frac {2- \frac{1}{n} } {2+\frac{1}{n}} | } = 1 $$ Therefore, the power series $Σ_{n=1}^∞ \frac{(-1)^{n-1}}{2n-1} x^{2n-1}$ is convergent for $|x| < 1$. Now for $ x = 1$ $Σ_{n=1}^∞ \frac{(-1)^{n-1}}{2n-1} = \frac{2}{1} - \frac {3}{3} + \frac{5}{5} - \frac{7}{7}+ ... $ **By Leibnitz test, it is convergent.** Since $\frac{1}{(2n+1)}$ is monotonic decreasing & $$lim_{n→∞} \frac{1}{(2n+1)} = 0$$ So, the power series $Σ_{n=1}^∞ \frac{(-1)^{n-1}}{2n-1}$ is convergent for $-1 < x ≤ 1$. ## Example: Expand tan⁻¹x in a power series. $Σ_{n=0}^∞ (x+1)^n = 1 - x + x^2 - x^3 + ....$ if $|x| < 1$ Integrating termwise we get: $$ tan⁻¹x = C + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... $$ if $|x|<1$ For $x = 0$, we get: $C = 0$ Therefore: $$ tan⁻¹x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^{n-1} \frac{x^{2n-1}}{2n-1} $$ ## Example: Show that: $$ π = 1 - \frac {1}{3}+ \frac {1}{5} - \frac {1}{7} + .... $$ (Gregory’s Series) **Solution:** $$ tan⁻¹(2x) = 2x -\frac{(2x)^3} {3} + \frac{(2x)^5}{5} -\frac{(2x)^7}{7} + ... $$ if $|x| < 1$ $$ \frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ... $$ ## Example: Prove that $ \frac{1}{2}(tan⁻¹x)² = \frac{x^2}{2} - \frac{x^4}{4} (1 + \frac {1}{3} ) + \frac{x^6}{6} (1 + \frac {1}{3} + \frac {1}{5} ) + ...$ if $|x| ≤ 1$ **Solution:** We have $$tan⁻¹x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... $$if $|x| ≤ 1$ and $$(1 + x^2) = 1 - x^2 + x^4 - x^6 + ... $$ if $|x| ≤ 1$ Both the series are ab. convergent $Σ_{n=0}^∞$, therefore their Cauchy product will converge absolutely to the product of their sums. $(1 + x^2). tan⁻¹x = (1 + x^2)(x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...) $ if |x| ≤ 1 Integrating termwise we get: $$\frac{1}{2}(tan⁻¹x)² = \frac{x^2}{2} - \frac{x^4}{4}(1 + \frac{1}{3}) + \frac{x^6}{6} (1 + \frac{1}{3} + \frac{1}{5}) + ... $$ if $|x|≤ 1$ The constant of integration vanishes. It can be easily seen that the power series on the right converges at $x=1$ also. So, that by Abel's theorem $$\frac{1}{2}(tan⁻¹x)² = \frac{x^2}{2} - \frac{x^4}{4} (1 + \frac{1}{3}) + \frac{x^6}{6}(1 + \frac{1}{3} + \frac{1}{5}) + ... $$ if $|x|≤1$ ## Example Show that: $log(1+x) = x - \frac{x^2}{2} + \frac {x^3}{3} - \frac{x^4}{4} + ... $ if $|x| ≤ 1$. and deduce that: $log2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... $ **Solution:** We know: $$(1 + x) = 1 - x + x^2 - x^3 + ... $$ if $|x| ≤ 1$ Integrating: $$log(1+x) = x - \frac {x^2}{2}+ \frac{x^3}{3} - \frac{x^4}{4} + ... $$ if $|x| ≤ 1$ The constant of integration vanishes by putting $x = 0$. The power series on the right converges at $x=1$ also. Therefore by Abel's theorem: $$log(1+x) = x - \frac {x^2}{2}+ \frac{x^3}{3} - \frac{x^4}{4} + ... $$ if $|x| ≤ 1$ ... $$ \frac{1}{2} [log(1+x)]² = \frac{x^2}{2} - \frac{(1+x)^4}{4} (1 + \frac{1}{3}) + \frac{(1+x)^6}{6} (1 + \frac {1}{3} + \frac{1}{5}) + ... $$ if $|x| ≤ 1$ **Solution:** We know: $$log(1 + x) = x - \frac{x^2}{2} + \frac {x^3}{3} - \frac{x^4}{4} + ... $$ if $|x| ≤ 1$ and $$(1 + x)^2 = 1 - x^2 + x^4 - x^6 + ... $$ if $|x| ≤ 1$ Both the series are ab. cgt, therefore their Cauchy product will converge to the product of their sums. $(1 + x)^2. log(1 + x) = (1 + x)^2 (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... + \frac{x^n}{n} + ... ) $ Integrating: $$ \frac{1}{2}log(1+x)²= \frac{x^2}{2} - \frac{(1+x)^4}{4}(1 + \frac {1}{3}) + \frac{(1+x)^6}{6}( 1 + \frac{1}{3} + \frac {1}{5}) + ... $$ if $|x| ≤ 1$ The constant of integration vanishes. Since the series on the right converge at 1. By Abel's theorem we have: $$ \frac{1}{2}log(1+x)²= \frac{x^2}{2} - \frac{(1+x)^4}{4}(1 + \frac {1}{3}) + \frac{(1+x)^6}{6}( 1 + \frac{1}{3} + \frac {1}{5}) + ... $$ if $|x| ≤ 1$ ## Abel’s Theorem **Statement:** Let $Σ_{n=0}^∞ C_nx^n $ be a power series with infinite radius of convergence R and (-1 < x < 1). Suppose $f(x) = Σ_{n=0}^∞ c_nx^n$ If $Σ_{n=0}^∞ c_nx^n$ converges then prove that: $lim_{x→1} f(x) = Σ_{n=0}^∞ c_n$ **Proof**: Let $Σ_{n=0}^∞ c_nx^n$ be a power series with unit radius of convergence and: $f(x) = Σ_{n=0}^∞ c_nx^n$ for$ (-1 ≤ x ≤ 1)$ Let $S_n = c_0 + c_1 + ... + c_n$. Let $Σ_{n=0}^∞ c_n = S$ then $S_{-1}=0$ Therefore: $lim_{n→∞} S_n = S$ Then: $$f(x) = (1-x)Σ_{n=0}^{m-1}(S_n-S_n-1)x^n$$ Let $m→∞$ since $Σ_{n=0}^{m-1}(S_n-S_n-1)x^n = Σ_{n=0}^{m-1} S_n x^n - Σ_{n=0}^{m-1} S_n - 1 x^n = 0$. For $|x| < 1$, when $m→∞$ since $m→∞$ and $S_{m} = S$ we get $f(x) = (1-x)Σ_{n=0}^∞ S_n x^n$ for $|x| <1$. Again, since $S_n → S$ for $ε>0$, $∃N$ such that: $|S_n - S| ≤ ε$ for $n ≥ N$ and $|x| < 1$ (2) Also: $|S_n - S| ≤ ε$ for $n ≥ N$, we have for $0 ≤ x ≤ 1$ (1) Hence: $$|f(x)-S| = |(1-x) Σ_{n=0}^{N-1}(S_n-S)x^n + (1-x)Σ_{n=N}^∞ (S_n-S)x^n|$$ $$ ≤ (1-x)|Σ_{n=0}^{N-1}(S_n-S)x^n| + (1-x)|Σ_{n=N}^∞ (S_n-S)x^n|$$ $$ ≤ (1-x)|Σ_{n=0}^{N-1}|S_n-S||x^n| + (1-x)|Σ_{n=N}^∞ |S_n-S||x^n| $$ $$ ≤ (1-x)Σ_{n=0}^{N-1}|S_n-S| + (1-x)Σ_{n=N}^∞ ε |x^n| $$ $$ ≤ (1-x)Σ_{n=0}^{N-1}|S_n-S| + ε * \frac {(1-x^{m+1})}{(1-x)} $$ $$ = (1-x)Σ_{n=0}^{N-1}|S_n-S| + ε $$ $$ = (1-x)Σ_{n=0}^{N-1}|S_n-S| + ε $$ $$ → 0$$ Therefore, $lim_{x→1}f(x) = S$. ## Tauber’s Theorem Let $f(x) = Σ_{n=0}^∞ c_nx^n$ (-1 < x < 1) If $f(x) → S$ as $x → 1$ then prove that $Σ_{n=0}^∞ c_n$ converges to S (i.e., lim $Σ_{n=0}^k c_n = S$ as $k → ∞$). **Proof:** To prove Tauber’s theorem, it is sufficient to prove that: $$lim_{k→∞} (Σ_{n=0}^k c_n x^n - Σ_{N=0}^k c_n) → 0$$ where $N = x$ for $|x|<1$. Given $ε > 0$ $∃ N$ so large that $| Σ_{n=0}^k c_n - S| < ε$. Name (Call) these two sums as $S_1$ and $S_2$ ($S_1 = Σ_{n=0}^k c_n x^n$ and $S_2 = Σ_{n=0}^N c_n$) Then: $|S_1| = |Σ_{n=0}^{N+1}c_nx^n| < ε | Σ_{n=0}^{N+1}c_n| < ε (Σ_{n=0}^{N+1} |c_n | < ε (Σ_{n=0}^{N+1} |c_n|) < ε$ $(Σ_{n=0}^{N+1} |c_n|) < ε$ Also, $|S_2| = |Σ_{n=0}^N c_n ( 1 - x^n)| < (N + 1) ε$ Therefore,