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I am trying to solve $\sum\limits_{k=1}^n\frac{2k+1}{k(k+1)}$ using summation by parts:

$\sum u\Delta v=uv-\sum Ev\Delta u$

$u = 2x+1, \Delta v=1/x(x+1)=(x-1)_{-2}$

$v=-(x-1)_{-1}=-1/x$

$\Delta u = 2, Ev = -1/(x+1)$

$\sum\frac{(2x+1)}{x(x+1)}\delta x=\frac{-2x-1}{x}+2\left(\sum x_{-1}\delta x\right)$

Is my solution correct up to this point? If it is, what is the anti-difference $\sum x_{-1}\delta x$? This seems to be the finite equivalent to $\int x^{-1}dx=ln(x)$.

Rodrigo de Azevedo
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dbyrne
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2 Answers2

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Your computations are correct, which is easy to see by differencing.

The anti-difference of $x_{-1} = \frac{1}{x+1}$ is known a Harmonic number: $$ \sum_x \frac{1}{x+1} = H_x $$ where $H_n = \sum_{k=1}^n \frac{1}{k}$.

Sasha
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I'm not sure this fully answers your questions, but it is easier to solve this problem without summation by part, simply by partial fraction expansion: $$ S_n = \sum_{k=1}^{n}\frac{1}{k} + \sum_{k=1}^{n}\frac{1}{k+1}=2H_n-1+\frac{1}{n+1}=2H_n-\frac{n}{n+1} $$ where $H_n$ is n'th harmonic number

Alex
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  • +1 for suggesting a more elegant method, but your answer is incorrect. The correct solution is: $2H_n-1+\frac1{n+1}=2H_n-\frac{n}{n+1}$. You miscalculated what happens at $k=n$. – dbyrne Mar 07 '13 at 01:12
  • @dbyrne: thanks, edited – Alex Mar 07 '13 at 04:22