# Thread: Root and Ratio Tests - Testing for divergence or convergence

1. ## Root and Ratio Tests - Testing for divergence or convergence

$\sum_{n=1}^{\infty}{\frac{(n+1)^{2n}}{2^nn^{2n}}}$

$\sum_{n=2}^{\infty}{({{{\frac{n}{n+1}}})^{n^{2}}}}$

I have tried both tests on the series, and both yield the limit going to 1 (test fails, may diverge or converge).

Is there a way to show either diverge or converge by another test, or comparison theorem? Can't seem to find anything! Will keep trying though!

Thank you!!

2. There is the integral test.

That could prove messy, though.

Integral test for convergence - Wikipedia, the free encyclopedia

3. Originally Posted by matt.qmar
$\sum_{n=1}^{\infty}{\frac{(n+1)^{2n}}{2^nn^{2n}}}$

$\sum_{n=2}^{\infty}{({{{\frac{n}{n+1}}})^{n^{2}}}}$

I have tried both tests on the series, and both yield the limit going to 1 (test fails, may diverge or converge).

Is there a way to show either diverge or converge by another test, or comparison theorem? Can't seem to find anything! Will keep trying though!

Thank you!!

I don't get it: in both series I get something very nice and less than one after applying the root test:

$\sqrt[n]{\frac{\left(n+1\right)^{2n}}{2^nn^{2n}}}=\frac{\l eft(n+1\right)^2}{2n^2} \xrightarrow [n \to \infty]{}\frac{1}{2}<1$

$\sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}=\left(\frac{n}{ n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^ n}\xrightarrow [n \to \infty]{}\frac{1}{e}$

Tonio

4. Originally Posted by tonio

$\sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}=\left(\frac{n}{ n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^ n}\xrightarrow [n \to \infty]{}\frac{1}{e}$
Why soes this tend to ${1 \over e}$ ?

5. Originally Posted by bigdoggy
Why soes this tend to ${1 \over e}$ ?

A very well know limit: $\left(1+\frac{1}{n}\right)^n \xrightarrow [n\to \infty] {}e$

Tonio