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January 23rd, 2016, 10:43 AM   #1
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Find the radius of convergence, and the interval of convergence, of the series.

Power Series:

Find the radius of convergence, and the interval of convergence of the series:

$\displaystyle \sum_{n=1}^{\infty }\frac{\sqrt{n}(2x+3)^{n}}{n^{2}+1}$

I have tried to solve this problem as following:

$\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}} \right |=\lim_{n\rightarrow \infty}\left | \frac{\sqrt{n+1}(2x+3)^{n+1}.(n^2+1)}{\sqrt{n}(2x+ 3)^n.((n+1)^2+1)} \right |= \left | 2x+3 \right | $

well, I have skipped some steps here because of too many codings.

so $\displaystyle -2< x< -1$


And:

$\displaystyle x=-1,
\sum_{n=1}^{\infty }\frac{\sqrt{n}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}}{n^2(1+\frac{1}{n^2})}=\frac{1}{n^{3/2}}$
Convergence, according to the root test

$\displaystyle x=-2,
\sum_{n=1}^{\infty }\frac{\sqrt{n}(-1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(-1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(-1)^n}{n^2(1+\frac{1}{n^2})}=\frac{(-1)^n}{n^{3/2}}$

Leibzig gives:
1: The term changes sign for each n
2: $\displaystyle a_{n+1}> a_{n}$
3: $\displaystyle \lim_{n\rightarrow \infty } a_{n}=0$

But my fussy teacher says that everything is ok except my final answer and it should be written in a correct order (mathematically). So could you please look and tell me what is wrong? thank you.

Last edited by skipjack; January 23rd, 2016 at 02:26 PM.
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January 23rd, 2016, 11:05 AM   #2
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$-2<x<-1$

did you check the endpoints of the interval for convergence?
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January 23rd, 2016, 02:34 PM   #3
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At x = -2, it is absolutely convergent (same expression as x = -1), therefore it is convergent.

Convergence at both ends gives interval $\displaystyle -2\leq x \leq-1$.
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