# Convergent/Divergent, find sum

1. Mar 25, 2013

### iRaid

1. The problem statement, all variables and given/known data
Determine whether the series is convergent or divergent. If it is convergent, find its sum.
$$\sum\limits_{n=1}^{\infty} (\frac{1}{e^n}+\frac{1}{n(n+1)})$$

2. Relevant equations

3. The attempt at a solution
So I found it's convergent:
$$\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{(n^{2}+n)})=\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})$$
$$\lim_{n \to \infty}((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})=0+0+0$$
∴ the sum is convergent.

Now how would I find the sum of the series?

2. Mar 25, 2013

### Dick

Having a limit of 0 doesn't show it's convergent. (1/e)^n is geometric. The other two terms from a telescoping series. Find the sums separately.

3. Mar 25, 2013

### iRaid

My professor said that a limit as n approaches infinity that equals zero means the series is convergent.. And I see what you're saying about the sum. Now I'm just confused on the convergent part.
Edit:
For the (1/e)^n I guess you could also say that since |1/e|<1, the series is convergent.

4. Mar 25, 2013

### Dick

Right for the (1/e)^n. And I hope that's not really what your professor said. If the limit is NOT zero, then it's NOT convergent. Approaching zero doesn't mean it converges. 1/n approaches zero but the series 1/n is not convergent.

5. Mar 25, 2013

### iRaid

Oh I see the problem here. 1/n is a harmonic series and divergent. Now how would I go about proving it is convergent?

6. Mar 25, 2013

### iRaid

Also I just checked me notes it says the following:
If $\sum_{n=1}^{\infty} a_{n}$ is convergent, then $\lim_{n \to \infty}a_{n}=0$ and the opposite if an does not equal 0.

I think there is a difference in what I said and what my notes say right?

7. Mar 25, 2013

### Dick

You know that the (1/e)^n part is convergent and you know it's sum. Now do the 1/n-1/(n+1) part. Like I said, it telescopes. If you know those two parts converge then the sum converges.

8. Mar 25, 2013

### Dick

Yes, what you quoted doesn't say that if the limit is 0 then the sum converges.

9. Mar 25, 2013

### iRaid

Ok then I did it correctly.
Thank you, again.

10. Mar 25, 2013

### iRaid

So you cannot say the opposite, if $\lim_{n \to \infty} a_{n}=0$ then $\sum_{n=1}^{\infty}a_{n}$ converges.?

11. Mar 25, 2013

### Dick

Doesn't the example of $a_n=1/n$ show that you can't?

12. Mar 25, 2013

### iRaid

I mean are there any other examples of this, other than the harmonic series?

13. Mar 25, 2013

### Dick

Many. 1/n^(1/2) doesn't converge. 1/log(n) doesn't converge. log(n)/n doesn't converge. But the limits are all zero.

14. Mar 25, 2013

### iRaid

Alright, thanks a lot.

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