Question

test the series for convergence or divergence

Answer 1

got some attention here !

Answer 2

\[\sum_{n=1}^{\infty}\frac{ n \cos n \pi }{ 2^{n} }\]

Answer 3

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{n\cos(\pi n) }{2^n}<\sum_{ n=1 }^{ \infty } ~ \frac{n^4 }{2^n}}\)

Answer 4

hint, cosine is never larger than 1 (or smaller than -1)

Answer 5

and that on the right will converge.

Answer 6

or that , should have just gone that way :)

Answer 7

where did the \(n^4\) come from? can't you just use \(n\)?

Answer 8

I am just giving an extent, that even that will convegre....

Answer 9

whatever, it is random, I can use n^{1.1} for that too

Answer 10

so It does converge then?

Answer 11

Yup

Answer 12

those terms go to zero lickety split
\(2^n\) grow much much (much) faster than \(n\)
compare for example \(10\) and \(2^{10}\) and 10 is a very small number