# Calculating series...

• Mar 2nd 2009, 08:49 AM
curious
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.

• Mar 2nd 2009, 09:51 AM
curious
Solved. It doesn't converge. :)
• Mar 2nd 2009, 10:10 AM
Jameson
Mind showing your solution so this thread may be useful to others later on? Otherwise it will be deleted. (Surprised)
• Mar 2nd 2009, 10:20 AM
curious
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.
• Mar 2nd 2009, 10:44 AM
Krizalid
Quote:

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.
• Mar 2nd 2009, 10:47 AM
Moo
Quote:

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. (Nod)

If $\lim a_k \neq 0$, then the series does not converge...
This is the contrapositive of $\sum a_k < + \infty \Rightarrow \lim a_k=0$
• Mar 2nd 2009, 10:49 AM
curious
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.