Question

I have the Series

\[\sum_{n=1}^{\infty} (1/e^{n} + 1/n(n+1))\]

How do I find the sum of this series as I know it is convergent because if I set
1/e^{n} + 1/n(n+1) = an
and take

lim an = 0
n->infinity

and since the limit of an as n approaches infinity is zero the series is convergent

So yeah How do I find the sum of convergent series.

Answer 1

Two things.
1. Just because \[\lim_{n\to\infty}a_n=0\]does not mean that the sum converges. For example, \[\sum_{n=1}^\infty \frac{1}{n}\]

2. split\[\sum_{n=1}^{\infty} (1/e^{n} + 1/n(n+1))=\sum_{n=1}^\infty\frac{1}{e^n}+\sum_{n=1}^\infty \frac{1}{n(n+1)}\]

Answer 2

first sum is geometric, with \(r=\frac{1}{e}\) second one is going to use partial fractions i bet
lets see what KingGeorge says maybe i am wrong

Answer 3

That's exactly what I was going to say ^^

Answer 4

@Australopithecus do you know how to compute these? first one is the easiest

Answer 5

I have one more thing to add however. Since your sum is from 1 to \(\infty\), you will need to subtract 1 from your geometric sum to account for the \(n=0\) term that you are not including.

Answer 6

yeah i always make a mistake on that one

Answer 7

If the


lim an = 0
n->infinity

rule does not hold why is it in my text book am I misunderstanding the theory in my text book

Answer 8

btw lets make sure it is clear why this conveges
\[\sum e^{-n}=\sum\frac{1}{e^n}\] converges because it is a geometric series with \(r=\frac{1}{e}<1\)

Answer 9

and
\[\sum\frac{1}{n(n+1)}\] converges because the denominator is a polynomial of degree 2, the numerator is a polynomial of degree 0, and \(2-0=2>1\)

Answer 10

right, I understand the first one, but don't you have to check

\[\sum_{\infty}^{n-1} 1/n(n+1)\]

Answer 11

to determine if the series is convergent?

Answer 12

as well

Answer 13

you check with your eyeballs using the method i wrote above

Answer 14

if you have a polynomial top and bottom and the denominator is greater in degree by more than one, it converges

Answer 15

No I understand that :)

Answer 16

do you know how to add it?

Answer 17

yes -_-

Answer 18

ok then you are in good shape right?

Answer 19

well I'm just confused by the theorem that involves taking the limit of an

Answer 20

Oh I get it now

Answer 21

I think you're confusing it with the theorem that says:

If \(\lim_{n\to\infty} a_n \neq0\) then the series diverges.

Note that this only tells you if it diverges, and nothing about convergence.

Answer 22

if we find that the limit of an = 0 then we know nothing about convergence or divergence but if we find anything else as the limit we know it is divergent

Answer 23

Precisely.