# Calculus I: (multi) infinite sums

• Mar 30th 2009, 02:59 AM
NonCommAlg
Calculus I: (multi) infinite sums
Another, well, interesting but not necessarily "challenging" problem from the series of problems that I started posting a few days ago. I hope you'll like it! (Smile)

Suppose the integer $\displaystyle n \geq 1$ is given. Evaluate $\displaystyle I_n=\sum_{j_1 = 1}^{\infty} \sum_{j_2 = 1}^{\infty} \cdots \sum_{j_n = 1}^{\infty} \frac{1}{j_1 j_2 \cdots j_n(j_1 + j_2 + \cdots + j_n)}.$
• Apr 1st 2009, 01:52 AM
bakerconspiracy
What is this, N Sums from 1 to infinity of $\displaystyle \frac{1}{j^{n}(nj)}=\frac{1}{nj^{n+1}}$?

What is the context of this problem? What are you trying to do with it; Since j and n are in N, this is a subseries of $\displaystyle \frac{1}{j}$ , so I believe this will diverge no matter what. (could be wrong here)

Almost looks like the end of an induction proof of sorts. Let me know if I can help you work on this, looks fun.
• Apr 1st 2009, 03:51 AM
mr fantastic
Quote:

Originally Posted by bakerconspiracy
What is this, N Sums from 1 to infinity of $\displaystyle \frac{1}{j^{n}(nj)}=\frac{1}{nj^{n+1}}$?

What is the context of this problem? What are you trying to do with it; Since j and n are in N, this is a subseries of $\displaystyle \frac{1}{j}$ , so I believe this will diverge no matter what. (could be wrong here)

Almost looks like the end of an induction proof of sorts. Let me know if I can help you work on this, looks fun.

Since this is the Challenging Problems and Puzzles subforum (where questions are posted as a challenge for other members rather than to get help with) I think you'll find that NonCommAlg already knows the solution to this question ....
• Apr 1st 2009, 08:37 AM
NonCommAlg
here's a hint: $\displaystyle \frac{1}{j_1 + j_2 + \cdots + j_n}=\int_0^1 x^{j_1 + j_2 + \cdots + j_n - 1} \ dx.$
• Apr 1st 2009, 10:40 AM
bobak
is it $\displaystyle \frac{\pi^{2n}}{6^n}$ ?

Bobak
• Apr 1st 2009, 12:58 PM
bkarpuz
Quote:

Originally Posted by bakerconspiracy
What is this, N Sums from 1 to infinity of $\displaystyle \frac{1}{j^{n}(nj)}=\frac{1}{nj^{n+1}}$?

What is the context of this problem? What are you trying to do with it; Since j and n are in N, this is a subseries of $\displaystyle \frac{1}{j}$ , so I believe this will diverge no matter what. (could be wrong here)

Almost looks like the end of an induction proof of sorts. Let me know if I can help you work on this, looks fun.

this is really funny (Rofl)
• Apr 1st 2009, 08:06 PM
NonCommAlg
Quote:

Originally Posted by bobak

is it $\displaystyle \frac{\pi^{2n}}{6^n}$ ?

Bobak

no it's not! the answer is $\displaystyle n! \zeta(n+1).$ (Evilgrin)

solution: starting with the hint i gave you:

$\displaystyle I_n=\int_0^1 \frac{1}{x} \sum_{j_1=1}^{\infty} \sum_{j_2=1}^{\infty} \cdots \sum_{j_n=1}^{\infty} \frac{x^{j_1 + j_2 + \cdots + j_n}}{j_1 j_2 \cdots j_n} \ dx=\int_0^1 \frac{1}{x} \sum_{j_1=1}^{\infty}\frac{x^{j_1}}{j_1} \sum_{j_2=1}^{\infty} \frac{x^{j_2}}{j_2} \cdots \sum_{j_n=1}^{\infty} \frac{x^{j_n}}{j_n} \ dx$

$\displaystyle =\int_0^1 \frac{1}{x} \left(\sum_{j=1}^{\infty} \frac{x^j}{j} \right)^n \ dx= \int_0^1 (-1)^n \frac{(\ln(1-x))^n}{x} \ dx=\int_0^{\infty} \frac{t^n e^{-t}}{1 - e^{-t}} \ dt$ [here we did the substitution: $\displaystyle \ln(1-x) = -t$]

$\displaystyle =\int_0^{\infty}t^n e^{-t} \sum_{k=0}^{\infty}e^{-kt} \ dt = \sum_{k=0}^{\infty} \int_0^{\infty}t^ne^{-(k+1)t} \ dt=\sum_{k=0}^{\infty}\frac{n!}{(k+1)^{n+1}}=n! \zeta(n+1).$ (Cool)
• Apr 1st 2009, 08:10 PM
o_O
Calc I ? What university did you go to (Surprised) Sheesh! lol.
• Apr 1st 2009, 08:21 PM
NonCommAlg
Quote:

Originally Posted by o_O

Calc I ? What university did you go to (Surprised) Sheesh! lol.

haha ... i agree the problem is not straightforward but you really don't need to know anything beyond Calc I (well, a good Calc I course i mean), right?
• Apr 1st 2009, 08:28 PM
o_O
Haha guess that says a lot about the education at the university I go to. At the very least, this would be Calc III material and that's without the recognition of the zeta function ! (Thinking)

Guess SFU is that far ahead, huh? ;)
• Apr 1st 2009, 08:40 PM
NonCommAlg
Quote:

Originally Posted by o_O

Guess SFU is that far ahead, huh? ;)

if you mean average calculus students, not at all! but there are some first year students in here who, with a little push, can easily solve this problem for at least n = 2.

by the way, i didn't do my undergraduate at SFU (or anywhere in Canada).(Smirk)
• Apr 1st 2009, 08:48 PM
o_O
Quote:

Originally Posted by NonCommAlg
by the way, i didn't do my undergraduate at SFU (or anywhere in Canada).(Smirk)

(Surprised) !?!?