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Taylor series (centered at -1) is given by:

$$ \sum_{n=1}^\infty \frac{(n+1)}{n}(x+1)^n $$

  1. what function centered at -1 does this series represent?

  2. hints as to how I may find its interval of convergence is (-2,0)?

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Thanks for editing the question, with appropriate symbols Matt. My first time here actually. –  student101 Sep 6 '12 at 17:03

2 Answers 2

up vote 3 down vote accepted

Hint: Let $w=1+x$. Note that $\dfrac{n+1}{n}=1+\dfrac{1}{n}$.

So our sum is $$\sum_1^\infty w^n +\sum_1^\infty \frac{1}{n}w^n.$$ The first sum will be very familiar. For the second, note that $\dfrac{w^n}{n}$ is an antiderivative of $w^{n-1}$.

For convergence, you are interested in showing that the interval is $-1\lt w\lt 1$. Ratio test will do it, except that you need to show also that we do not have convergence at $w=\pm 1$.

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Left term in sum: {1/(1-w)} - 1 –  student101 Sep 6 '12 at 17:42
    
@student101: Indeed, and now you should be ready for right term. –  André Nicolas Sep 6 '12 at 17:46
    
right term in sum: {1/(1-w)}d/dw= 1/{(1-w)^2} –  student101 Sep 6 '12 at 17:46
    
@student101: Integrate, don't differentiate. Answer looks like $\int_0^w f(t)\,dt$ for suitable $f$. –  André Nicolas Sep 6 '12 at 17:52
    
integral being -log(1-w) –  student101 Sep 6 '12 at 17:53

For the interval of convergence, consider the root test, and answer the following questions:

What is $\lim_{n \to \infty} (\frac{n+1}{n})^{1/n}$?

For what values of $x$ is $|x+1| < 1$?

For what values of $x$ is $|x+1| = 1$?

When $|x+1| = 1$, does the series converge or diverge?

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so (1+1/n)^(1/n) converges to 1 –  student101 Sep 6 '12 at 17:33
    
abs(x+1)<1 for -2<x<0 –  student101 Sep 6 '12 at 17:35
    
at x=0, the series converges; thanks :) –  student101 Sep 6 '12 at 17:37
    
at x=-2, series converges also –  student101 Sep 6 '12 at 18:06
    
How can it converge at $|x+1| = 1$ when the absolute value of each tern is greater than $1$? –  marty cohen Sep 7 '12 at 17:29

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