# Sum of a finite series.

• Dec 11th 2012, 02:20 AM
Kaloda
Sum of a finite series.
Show:

1/2 + 1/6 + 1/12 + 1/20 + .... = 1
• Dec 11th 2012, 02:45 AM
MarkFL
Re: Sum of a finite series.
The result you cite is true for the infinite series:

$\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$

To find the partial sum, we may write (using partial fraction decomposition):

$S_n=\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n} \frac{1}{k}-\sum_{k=1}^{n} \frac{1}{k+1}=\sum_{k=1}^{n} \frac{1}{k}-\sum_{k=2}^{n+1} \frac{1}{k}=?$

Can you simplify this by observing all but two terms cancel each other out and observe what happens as $n\to\infty$?
• Dec 11th 2012, 04:28 AM
Kaloda
Re: Sum of a finite series.
Oh yeah.

Btw, that problem is just a part of my long solution for this main problem:
Find the sum of <tex>sqrt(1+\frac{1}{1^2}+\frac{1}{2^2})+sqrt(1+\f rac{1}{2^2}+\frac{1}{3^2})+sqrt(1+\frac{1}{3^2}+\f rac{1}{4^2})+...+sqrt(1+\frac{1}{1999^2}+\frac{1}{ 2000^2})</tex>.

I aproached this main problem by observing the simplified form of each first 5 terms and I thought that the sum is just equal to the number of terms to the aforementioned series. But now that you mentioned that solution, then maybe there's another or perhaps general solution for this problem without using logic or any patterns.

I know that the series above can be expressed as the sum of <tex>1+\frac{1}{k^2}+\frac{1}{(k+1)^2}</tex> with n=1 to n=1999.
• Dec 11th 2012, 04:54 AM
MarkFL
Re: Sum of a finite series.
We are given to evaluate:

$\sqrt{1+ \frac{1}{1^2}+ \frac{1}{2^2}}+\sqrt{1+ \frac{1}{2^2}+ \frac{1}{3^2}}+\sqrt{1+ \frac{1}{3^2}+ \frac{1}{4^2}}+\cdots+\sqrt{1+ \frac{1}{1999^2}+ \frac{1}{ 2000^2}}$

Let's write this as:

$S_{1999}=\sum_{k=1}^{1999}\sqrt{1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2}}$

and so:

$S_{n}=\sum_{k=1}^{n}\sqrt{1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2}}$

Now, we may rewrite this as:

$S_{n}=\sum_{k=1}^{n} \frac{k^2+k+1}{k(k+1)}$

Using division and partial fraction decomposition, this becomes:

$S_{n}=\sum_{k=1}^{n}\left(1+ \frac{1}{k}- \frac{1}{k+1} \right)$

Can you proceed from here, using a similar technique as above?
• Dec 11th 2012, 05:13 AM
Kaloda
Re: Sum of a finite series.
Thank you so muchhhhhh.

But here's my last question. How can I show that k^2*(k+1)^2+k^2+(k+1)^2 is equal to (k^2+k+1)^2. I know that the former one is equal to k^4+2k^3+3k^2+2k+1 but i can't seem to factor it to form the latter.
• Dec 11th 2012, 05:25 AM
MarkFL
Re: Sum of a finite series.
Here is the technique I used:

$1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2}=$

$\frac{(k(k+1))^2+(k+1)^2+k^2}{(k(k+1))^2}$

After expanding and collecting like terms, the numerator becomes:

$k^4+2k^3+3k^2+2k+1$

Now, we wish to see if it may be factored as the square of a quadratic, and we know what the leading and trailing terms must be, so we may write:

$k^4+2k^3+3k^2+2k+1=(k^2+ak+1)^2$

Expanding the right side, we find:

$k^4+2k^3+3k^2+2k+1=k^4+2ak^3+(a^2+2)k^2+2ak+1$

This implies that $a=1$, and so we have:

$k^4+2k^3+3k^2+2k+1=(k^2+k+1)^2$

Hence:

$1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2}=\left( \frac{k^2+k+1}{k(k+1)} \right)^2$