How do we evaluate this sum: \sum_{i=1}^{10}\frac{2i+1}{i^2(i+1)^2}?

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SUMMARY

The sum \(\sum_{i=1}^{10}\frac{2i+1}{i^2(i+1)^2}\) evaluates to a specific numerical value through the application of partial fraction decomposition and simplification techniques. The correct solutions were provided by members including hxthanh and Chris L T521, showcasing a collaborative effort in problem-solving. The discussion emphasizes the importance of understanding series and sequences in mathematical evaluations.

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  • Understanding of partial fraction decomposition
  • Familiarity with summation notation
  • Knowledge of series convergence
  • Basic algebraic manipulation skills
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  • Learn about convergence tests for series
  • Explore advanced summation techniques and identities
  • Practice evaluating similar sums with different limits and functions
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Mathematics students, educators, and anyone interested in advanced summation techniques and series evaluations will benefit from this discussion.

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Evaluate the following sum:

$$\sum_{i=1}^{10}\frac{2i+1}{i^2(i+1)^2}$$
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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):
\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: $S_{10}=1-\dfrac{1}{121}=\boxed{\dfrac{120}{121}}$
 

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