(a) Differentiate the Maclaurin series for 1/(1 x),and use the result to show thatk=1kxk

Chapter 9, Problem 36

(choose chapter or problem)

(a) Differentiate the Maclaurin series for $1 /(1-x)$, and use the result to show that

$\sum_{k=1}^{\infty} k x^{k}=\frac{x}{(1-x)^{2}} \text { for }-1<x<1$

(b) Integrate the Maclaurin series for $1 /(1-x)$, and use the result to show that

$\sum_{i=1}^{\infty} \frac{x^{k}}{k}=\ln (1-x) \text { for }-1<x<1$

$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{\alpha^{k}}{k}=\ln (1+x) \text { for }-1<x<1$

(d) Show that the series in part (c) converges if $x=1$.
(e) Use the remark following Example 3 to show that

$\sum_{k=1}^{\infty}(-\mid 1)^{k+1} \frac{x^{k}}{k}=\ln (1+x) \text { for }-1<x \leq 1$

Equation Transcription:

$1/(1-x)$

$x=1$

$\sum\limits_{k=1}^{\infty{}}k{x}^{k}=\frac{x}{(1-x{)}^{2}}\ for\ -1\ <x<1$

$\sum\limits_{k=1}^{\infty{}}(-1{)}^{k+1}=In(1-x)\ \ for\ -1\ <x<1$

$\sum\limits_{k=1}^{\infty{}}(-1{)}^{k+1}\frac{{x}^{k}}{k}=In(1+x)\ \ for\ -1\ <x<1$

$\sum\limits_{k=1}^{\infty{}}(-1{)}^{k+1}\frac{{x}^{k}}{k}=In(1+x)\ \ for\ -1\ <x\leq{}1$

Text Transcription:

1/(1-x)

x=1

Sum of k=1 infinity kx^k = x/(1-x)^2 for -1 < x < 1

Sum of k=1 infinity (-1)^k+1 = In(1-x) for -1 < x <1

Sum of k=1 infinity (-1)^k+1 x^k/k= In(1+x) for -1 < x < 1

Sum of k=1 infinity (-1)^k+1 x^k/k = In(1+x) for -1 < x leq 1

Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.

Login