# Math Help - Calculating series...

1. ## Calculating series...

Hey, folks!

How would you calculate the sum of (k^(j + n))/(m + k)^n, where summation variable is k and k = 1 to infinity?

j, n, and m are positive integers.

2. Solved. It doesn't converge.

3. Mind showing your solution so this thread may be useful to others later on? Otherwise it will be deleted.

4. Well, the necessary condition for series to converge is lim a_k -> 0 as k tends to infinity.

In my case, a_k = (k^(j + n))/(m + k)^n = (k^j)*(1 - m/(m + k))^n.

As k -> infinity, (1 - m/(m + k))^n -> 1, while k^j -> infinity.

Which means a_k is unbounded. So, the necessary condition is not satisfied.

5. Originally Posted by curious

Well, the necessary condition for series to converge is lim a_k -> 0 as k tends to infinity.
Consider $\sum_{n\ge1}\frac1n,$ whereas $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0,$ but the series diverges anyway.

6. Originally Posted by Krizalid
Consider $\sum_{n\ge1}\frac1n,$ whereas $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0,$ but the series diverges anyway.
He said necessary, not sufficient, so what he did is nothing else but correct.

If $\lim a_k \neq 0$, then the series does not converge...
This is the contrapositive of $\sum a_k < + \infty \Rightarrow \lim a_k=0$

7. Yeah, sure. It's only the necessary condition.

That is, if a_k -> 0, it doesn't mean that the series will converge.

But if a_k -> "something other than 0", then definitely the series won't converge.