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Find the value of

\(\frac{2\times 1 +1}{1^2 \times 2^2}+\frac{2\times 2 +1}{2^2 \times 3^2}+\frac{2\times 3 +1}{3^2\times 4^2}+...\)

 May 24, 2020
 #1
avatar+995 
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I am unsure at what term this sequence ends.

 May 24, 2020
 #2
avatar+146 
+1

that can be rewritten as this:
\(\sum_{n=1}^{\infty}\frac{2n+1}{n^2(n+1)^2}\)

 

so first we have to find the formula for the nth partial sum

\((1)\sum_{n=1}^{\infty}\frac{2n+1}{n^2(n+1)^2}\\ (2)\sum_{n=1}^{\infty}\frac{2n+1+n^2-n^2}{n^2(n+1)^2}\\ (3)\sum_{n=1}^{\infty}\frac{n^2+2n+1}{n^2(n+1)^2}-\frac{n^2}{n^2(n+1)^2}\\ (4)\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{1}{(n+1)^2}=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...+\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)=1-\frac{1}{(n+1)^2}\\\)

step 2: try to turn the series into a more recognizable telescoping series 

step 3: split the fraction

step 4: the terms cancel out leaving 1-1/(n+1)^2

 

now we take the limit to infinity of the nth partial sum formula to find the value

\(\lim_{n\rightarrow \infty}1-\frac{1}{(n+1)^2} = 1\)

since you're in algebra another way to look at the thing above is as n gets bigger and bigger to infinity, 1/(n+1)^2 will become 0, leaving 1-0 = 1

 May 24, 2020

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