Method of telescopic sums....help

**Craige** · July 19th 2008, 03:05 AM

hi all i need to solve,

Using method of telescopics sums to determine that;

999
Σ 2k+1/k^2(k+1)^2 = 0.000099
k=100

Please help!!!!

**flyingsquirrel** · July 19th 2008, 03:34 AM

Hi

Originally Posted by Craige

hi all i need to solve,

Using method of telescopics sums to determine that;

999
Σ 2k+1/k^2(k+1)^2 = 0.000099
k=100

As you did not use enough parenthesis, I guess we should understand $\sum_{100}^{999}\frac{2k+1}{k^2(k+1)^2}$ . Then note that $\frac{1}{k^2}-\frac{1}{(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}$ . (check it !) Using this, you should be able to conclude. Good luck !

**Krizalid** · July 19th 2008, 07:07 AM

Originally Posted by flyingsquirrel

Then note that $\frac{1}{k^2}-\frac{1}{(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}$ . (check it !)

The thing is, not all people is able to see this. Of course, pretty easy when given the partial fraction descomposition and then just do algebra to verify it.

To make what flyingsquirrel did, first note that $2k+1=(k^2+2k+1)-k^2,$ nothing weird, we just added & subtracted a convenient term. Following up on this, we have that $2k+1=(k+1)^2-k^2.$ Now, what's the cool thing here: the cool here tell us that in the denominator we have exactly the terms $k^2$ & $(k+1)^2.$ Hence, $\frac{2k+1}{k^2(k+1)^2}=\frac{(k+1)^2-k^2}{k^2(k+1)^2}.$

Finally, split this original ratio into two parts, like $\frac{x+y}{z}=\frac xz+\frac yz,$ and the last ratio becomes $\frac{(k+1)^2}{k^2(k+1)^2}-\frac{k^2}{k^2(k+1)^2}.$ Then after makin' simple simplifications, we happily get that $\frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}.$

-----

Back to the sum and let $a_k=\frac1{k^2}\implies a_{k+1}=\frac1{(k+1)^2}.$ So we can see the sum as $\sum_{k=100}^{900}(a_k-a_{k+1}).$ Add some terms: $(a_{100}-a_{101})+(a_{101}-a_{102})$ and so on. Can you see what's goin' on here?

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