• Nov 14th 2009, 06:58 AM
usagi_killer
1. Find a formula for the sum http://alt2.artofproblemsolving.com/...894c1b290f.gif

2. Find the sum of http://alt2.artofproblemsolving.com/...5c62244619.gif

3. Find a formula for http://alt2.artofproblemsolving.com/...8126ed417e.gif

Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

Thanks!
• Nov 14th 2009, 07:38 AM
galactus
For #1: It can be rewritten as $\sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$

The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

The second one $\sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2$

$e-1-(e-2)=1$
• Nov 14th 2009, 09:29 AM
tonio
Quote:

Originally Posted by usagi_killer
1. Find a formula for the sum http://alt2.artofproblemsolving.com/...894c1b290f.gif

2. Find the sum of http://alt2.artofproblemsolving.com/...5c62244619.gif

$\log_ku=\frac{\ln u}{\ln k}\Longrightarrow \sum\limits_{k=2}^n\frac{1}{\log_ku}=$ $\frac{1}{\ln u}\sum\limits_{k=2}^n\ln k=\frac{1}{\ln u}\ln\left(\prod\limits_{k=2}^nk\right)=\frac{\ln k!}{\ln u}=\log_uk!$

Tonio

3. Find a formula for http://alt2.artofproblemsolving.com/...8126ed417e.gif

Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

Thanks!

.
• Nov 14th 2009, 09:54 AM
Krizalid
Quote:

Originally Posted by usagi_killer

$\sum\limits_{j=1}^{n}{\frac{1}{j(j+1)(j+2)}}=\frac {1}{2}\sum\limits_{j=1}^{n}{\frac{(j+2)-j}{j(j+1)(j+2)}},$ hence your sum equals $\frac{1}{2}\sum\limits_{j=1}^{n}{\left( \frac{1}{j(j+1)}-\frac{1}{(j+1)(j+2)} \right)}=\frac{1}{2}\left( \frac{1}{2}-\frac{1}{(n+1)(n+2)} \right).$
• Nov 14th 2009, 07:26 PM
usagi_killer
Quote:

Originally Posted by galactus
For #1: It can be rewritten as $\sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$

The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

The second one $\sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2$

$e-1-(e-2)=1$

Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?
• Nov 15th 2009, 01:01 AM
tonio
Quote:

Originally Posted by usagi_killer
Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?

Galactus did a nice job...but with an infinite series, whereas you wanted a finite one, yet the answer STILL is in Galactus message:

$\sum\limits_{k=1}^n\frac{k}{(k+1)!}=\sum\limits_{k =1}^n\frac{1}{k!}-\sum\limits_{k=1}^n\frac{1}{(k+1)!}=1-\frac{1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$

Tonio