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Find the interval of convergence including endpoints of $$\sum_{n=1}^{\infty} \frac{n(x+3)^n}{2^n(n^2+1)}.$$ I can find the interval of convergence but I don't know how to test endpoints $x=-5$ and $x=-1$. I don't know whether the series convergences at $x$ = $-5$ and $x$=$-1$ and need help. I don't think the series converges at $x= -5$ and $x= -1$.

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  • $\begingroup$ Since you are new @MSE, I think its time for you to read this thread $\endgroup$ Mar 26, 2017 at 1:31

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At $x=-5$, the series becomes $$\sum_{n=1}^{\infty}\frac{n(x+3)^n}{2^n(n^2+1)} =\sum_{n=1}^{\infty}\frac{n(-2)^n}{2^n(n^2+1)} =\sum_{n=1}^{\infty}\frac{(-1)^nn}{n^2+1}.$$

Let $a_n=\frac{n}{n^2+1}$, where $n\in\Bbb N$. The $$\begin{align} \frac{a_{n+1}}{a_n}&=\frac{n+1}{n^2+2n+2}\cdot\frac{n^2+1}{n}\\ &=\frac{n^3+n^2+n+1}{n^3+2n^2+2n}\\ &=\frac{n^3+n^2+n+1}{n^3+n^2+n+n^2+n}<1.\\ \end{align}$$ Next, $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n}{n^2+1}=0.$$ So, the series $$\sum_{n=1}^{\infty}\frac{(-1)^nn}{n^2+1}$$ converges (by Alternating Series Test).

At $x=-1$, the series becomes $$\sum_{n=1}^{\infty}\frac{n(x+3)^n}{2^n(n^2+1)} =\sum_{n=1}^{\infty}\frac{n(2)^n}{2^n(n^2+1)}=\sum_{n=1}^{\infty}\frac{n}{n^2+1}.$$

For each $n\in\Bbb N$, we have $$\frac{n}{n^2+1}\geq \frac{n}{n^2+n^2}=\frac{n}{2n^2}=\frac{1}{2}\cdot\frac{1}{n}.$$ We know that $$\sum_{n=1}^{\infty}\bigg(\frac{1}{2}\cdot\frac{1}{n}\bigg)$$ diverges and so, the series $$\sum_{n=1}^{\infty}\frac{n}{n^2+1}$$ diverges (by Comparison Test).

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  • $\begingroup$ Actually, the index of the series is is 0 not 1. $\endgroup$
    – user420360
    Mar 22, 2017 at 0:52
  • $\begingroup$ @brittany Well, at $n=0$, the first term of the series is zero. So, what's the use of starting the index at $n=0$? $\endgroup$ Mar 22, 2017 at 3:42

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