Question

determine whether the following series converges or diverges
∑_((n+4^n)/((n+6^n)) fm n=2to infinity

Answer 1

u can use limit cmparisn test or compaisn test

Answer 2

\[ \left(\frac{4}{6} \right )^n\]

Answer 3

I can't see the sequence properly!!!

Answer 4

it is he series from 2 to nfinity plz check i on the atatchment

Answer 5

wolframalpha says it converges
http://www.wolframalpha.com/input/?i=sum%28%28n%2B4^n%29%2F%28n%2B6^n%29%2C+2%2Cinfinity%29

Answer 6

Limit test works
http://www.wolframalpha.com/input/?i=lim+n-%3Einfinity+%28%28n%2B1%2B4^%28n%2B1%29%29%2F%28n%2B1%2B6^%28n%2B1%29%29%29%2F%28%28n%2B4^n%29%2F%28n%2B6^n%29%29

Answer 7

@REMAINDER sorry I meant I have a proof it converges :D

Answer 8

\[ \frac{(n) + (4)^n}{n+6^n} = \frac{(n)}{n+6^n} + \frac{(4)^n}{n+6^n} \]
And \( \frac{4^n}{n + 6^n} < (4/6)^n\) so it converges

The other part \( \frac{(n)}{n+6^n} < \frac{n}{6^n}\) which can be shown converge with simple ratio test.

Hence it converges!!!

Answer 9

So here it is:
\[\sum_{n=2}^{\infty}(n+4^n)/(n+6^n)=\sum_{n=2}^{\infty}(n/n+6^n)+(4^n/n+6^n)\]
\[=\sum_{n=2}^{\infty}(n/n+6^n)+\sum_{n=2}^{\infty}(4^n/n+6^n)\]
so if these two series are convergent, then our series, which is a sum of two convergent series, will ofcourse be convergent. So all we need to prove now is that the two are in fact convergent.
let's first start with the first one:
so for this one I'll use the divergence test. we know that if:
\[\lim_{n \rightarrow \infty}n/n+6^n=0\]
the first series will be convergent. However, because of that naughty little n, i'm going to use L' Hopital's Rule so:
\[ \lim_{n \rightarrow \infty}n/n+6^n=\lim_{n \rightarrow \infty}1/1+6^nln6=0\]
by divergence test, the first series will converge
now for the second one:
I'll use comparison test.
as @experiment said correctly,
4^n/n+6^n<(4/6)^n
so since the second term is less than a geometric series with ratio 4/6, hence convergent, then second term will also be converget.
and since the series we are after is just the sum of two convergent series, it will also be convergent..