# Pick out the convergent series

Pick out the convergent series:

1. $\sum_{n=1}^\infty(n + 1)^n/n^{n+3/2}$
2. $1 +(1/2^2) +(2^2/3^3 )+(3^3/4^4) + \cdots$
3. $\sum_{n=1}^\infty\sqrt {(1 + 4^n) /(1 + 5^n)}$ .

totally stuck on this problem.how can I able to solve this problem.thanks for your time.

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Have you tried rewriting any of the summands? Try putting things in a more familiar form. Also, is this homework? If so, please add the homework tag. –  Antonio Vargas Jan 14 '13 at 5:15

(a) In looking at our expression, the eye I think goes to $\frac{(n+1)^n}{n^n}$, because it is $\left(1+\frac{1}{n}\right)^n$, whose behaviour we are familiar with. More formally, $$\frac{(n+1)^n}{n^{n+3/2}}=\left(1+\frac{1}{n}\right)^n \frac{1}{n^{3/2}}.$$ We know that $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$. In particular, if $n$ is large enough, $\left(1+\frac{1}{n}\right)^n \lt 3$. (Indeed, the expression is always $\lt 3$, but we don't need to know that.) So if $n$ is large enough, $$\frac{(n+1)^n}{n^{n+3/2}}\lt \frac{1}{n^{3/2}}.$$ But $\sum_{1}^\infty \frac{1}{n^{3/2}}$ converges, so by Comparison our series converges.

(b) We are looking at the series $\sum_0^\infty \frac{n^n}{(n+1)^{n+1}}$. One approach is much like the one we took in (a). Note that the $n$-th term is equal to $\frac{n^n}{(n+1)^n (n+1)}$. If we divide top and bottom by $n^n$, we find that the $n$-th term is equal to $$\frac{1}{\left(1+\frac{1}{n}\right)^n}\frac{1}{n+1}.$$ After a while (indeed always), $\left(1+\frac{1}{n}\right)^n \lt 3$, and therefore $$\frac{1}{\left(1+\frac{1}{n}\right)^n}\frac{1}{n+1}\gt \frac{1}{3}\frac{1}{n+1}.$$ The series $\sum_0^\infty \frac{1}{n+1}$ is the familiar harmonic series, which diverges. So by comparison our series diverges.

(c) The idea is that the series more or less behaves like $\sum_1^\infty\left(\left(\frac{4}{5}\right)^{1/2}\right)^n$, a convergent geometric series, because $(4/5)^{1/2}\lt 1$.

For detail, note that $1+4^n \lt 2\cdot 4^n$, while $1+5^n \gt 5^n$. It follows that $$\sqrt{(1+4^n)/(1+5^n)}\lt \sqrt{2}\left(\left(\frac{4}{5}\right)^{1/2} \right)^n.$$ But the series $$\sum_1^\infty \left(\left(\frac{4}{5}\right)^{1/2} \right)^n$$ converges, so by comparison our series also does.

Alternately, we could use the Ratio Test, or the Root Test.

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All the series converge, except the second one

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please tell me how can I solve it –  asutos Jan 14 '13 at 5:33