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!!!!
Hi
As you did not use enough parenthesis, I guess we should understand $\displaystyle \sum_{100}^{999}\frac{2k+1}{k^2(k+1)^2}$. Then note that $\displaystyle \frac{1}{k^2}-\frac{1}{(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}$. (check it !) Using this, you should be able to conclude. Good luck !
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 $\displaystyle 2k+1=(k^2+2k+1)-k^2,$ nothing weird, we just added & subtracted a convenient term. Following up on this, we have that $\displaystyle 2k+1=(k+1)^2-k^2.$ Now, what's the cool thing here: the cool here tell us that in the denominator we have exactly the terms $\displaystyle k^2$ & $\displaystyle (k+1)^2.$ Hence, $\displaystyle \frac{2k+1}{k^2(k+1)^2}=\frac{(k+1)^2-k^2}{k^2(k+1)^2}.$
Finally, split this original ratio into two parts, like $\displaystyle \frac{x+y}{z}=\frac xz+\frac yz,$ and the last ratio becomes $\displaystyle \frac{(k+1)^2}{k^2(k+1)^2}-\frac{k^2}{k^2(k+1)^2}.$ Then after makin' simple simplifications, we happily get that $\displaystyle \frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}.$
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Back to the sum and let $\displaystyle a_k=\frac1{k^2}\implies a_{k+1}=\frac1{(k+1)^2}.$ So we can see the sum as $\displaystyle \sum_{k=100}^{900}(a_k-a_{k+1}).$ Add some terms: $\displaystyle (a_{100}-a_{101})+(a_{101}-a_{102})$ and so on. Can you see what's goin' on here?