# Homework Help: Deciding if the series converges or diverges

1. Jan 2, 2010

### Lord Dark

1. The problem statement, all variables and given/known data
Hi everyone ,, got the following question

Sum(k=1,infinity) (2009+5^(-k^3))/(k^2(1+e^(-k^2)))

2. Relevant equations

3. The attempt at a solution

first I think it diverges so I did a comparison which is :
ak = (2009+5^(-k^3))/(k^2(1+e^(-k^2)))
e^(k^2))/(5^(k^3)*k^2) which is less than ak ,,

but now what should I do ?? which test should I use to make it diverges (that is if it diverges)

note : I tried already ratio test and it didn't work and k-th term but I get (infinity over infinity) so it's no use ,, can someone help please

2. Jan 2, 2010

### HallsofIvy

It should be easy to see that
$$\frac{2009+ 5^{-k^3}}{1+ e^{-k^2}}< 2010$$
compare this series to $\sum 2010/k^2$.

Thanks, jgens, for catching the typo.

Last edited by a moderator: Jan 3, 2010
3. Jan 2, 2010

### Lord Dark

lol ,, first of all ,, can you explain it a little further ,, how can you choose a number where the function is less than it ( i mean 20010) how did you know that number ??

and why did you remove 5^(-k^3) and e^(-k^2) ?? do they cancel each other or what ?

4. Jan 2, 2010

### jgens

Halls mistyped and probably meant that

$$\frac{2009 + 5^{-k^3}}{1 + e^{-k^2}} < 2010$$

This inequality should be obvious since $5^{-k^3} < 1$ which implies that $2009 + 5^{-k^3} < 2010$. Now, because $1 + e^{-k^2} > 1$ we know that

$$\frac{2009 + 5^{-k^3}}{1 + e^{-k^2}} < 2010$$

Now, use the comparrison test as Halls suggested.

5. Jan 3, 2010

### Lord Dark

aha ,, now I got you ... thanks :)