Jameson Insights Author Gold Member MHB Messages 4,533 Reaction score 13 Thread starter Sep 22, 2013 #1 Evaluate the following sum: $$\sum_{i=1}^{10}\frac{2i+1}{i^2(i+1)^2}$$ --------------------
Jameson Insights Author Gold Member MHB Messages 4,533 Reaction score 13 Sep 29, 2013 #2 Congratulations to the following members for their correct solutions: 1) hxthanh 2) Chris L T521 3) MarkFL 4) eddybob123 5) johng 6) anemone Solution (from hxthanh): Spoiler \begin{align*}S_n=\sum_{i=1}^n \dfrac{2i+1}{i^2(i+1)^2}&=\sum_{i=1}^n \dfrac{(i+1)^2-i^2}{i^2(i+1)^2}\\&=\sum_{i=1}^n \left(\dfrac{1}{i^2}-\dfrac{1}{(i+1)^2}\right)\\&=\dfrac{1}{1^2}-\dfrac{1}{(n+1)^2}\end{align*} With $n=10$, we get: $S_{10}=1-\dfrac{1}{121}=\boxed{\dfrac{120}{121}}$
Congratulations to the following members for their correct solutions: 1) hxthanh 2) Chris L T521 3) MarkFL 4) eddybob123 5) johng 6) anemone Solution (from hxthanh): Spoiler \begin{align*}S_n=\sum_{i=1}^n \dfrac{2i+1}{i^2(i+1)^2}&=\sum_{i=1}^n \dfrac{(i+1)^2-i^2}{i^2(i+1)^2}\\&=\sum_{i=1}^n \left(\dfrac{1}{i^2}-\dfrac{1}{(i+1)^2}\right)\\&=\dfrac{1}{1^2}-\dfrac{1}{(n+1)^2}\end{align*} With $n=10$, we get: $S_{10}=1-\dfrac{1}{121}=\boxed{\dfrac{120}{121}}$