Sum of a finite series.

**Kaloda** · December 11th 2012, 02:20 AM

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1/2 + 1/6 + 1/12 + 1/20 + .... = 1

**MarkFL** · December 11th 2012, 02:45 AM

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 $n\to\infty$ ?

**Kaloda** · December 11th 2012, 04:28 AM

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.
But how can I show this? Please help.

**MarkFL** · December 11th 2012, 04:54 AM

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?

**Kaloda** · December 11th 2012, 05:13 AM

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.

**MarkFL** · December 11th 2012, 05:25 AM

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:

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

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