Question

If I have the following sequence

Answer 1

\[a _{n} = \frac{ n }{ 2^{n} }\]

Answer 2

How can I determine that
\[\sum_{n=1}^{\infty} a _{n}\] converges or diverges

Answer 3

Ratio and Root tests are both inconclusive

Answer 4

and integral test is too complicated

Answer 5

Hmm... Comparison test maybe? Just gotta find a larger function which converges

Answer 6

I was thinking if I can prove that it's a Cauchy sequence

Answer 7

Are you sure the integral test is too complex?

Change the 2^x into e^(xln2) then you can prob integrate by parts.

Answer 8

what do I choose as my u and dv?

Answer 9

\[\large \int\limits_{1}^{\infty} \frac{ x }{ e ^{x \ln 2} }dx = \int\limits_{1}^{\infty} x e ^{-x \ln 2} dx \]

Answer 10

u = x so du = dx. \[dv = e ^{-x \ln 2}\] so \[v =\frac{ -e ^{-x \ln 2} }{ \ln2 }\]

Answer 11

okay I'll try that

Answer 12

It should converge, just off the fact that the 2^n grows much faster than the n, in n/2^n.

Answer 13

\[ |S_n - S_m| = \left | \frac{m}{2^m } + \frac{m+1}{2^{m+1} } + ... + \frac{n }{2^n }\right |\]