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How to calculate the limit of this sum? $$\lim _{n\to +\infty }\left(\sum _{k=n}^{2n}\:\frac{1}{k\left(k+1\right)}\right)\:$$

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    $\begingroup$ $\frac 1{k(k+1)} = \frac 1 k - \frac 1 {k+1}$. The RHS of your equation is nonsense. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 15:11
  • $\begingroup$ Welcome to math stack exchange. Please explain how you attempted to solve the problem. What do you know so far about limits and summations? What exact step are you stuck on? How did you stumble on this problem? $\endgroup$ – Arbuja Mar 6 '16 at 15:12
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Note that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Thus we have:

$$ \sum_{k = 1} ^ n \frac{1}{k(k+1)} = \sum_{k = 1} ^ n \frac{1}{k} - \sum_{k = 1} ^ n \frac{1}{k+1} = 1 - \frac{1}{n+1} $$

At last:

$$ \sum _{k=n}^{2n}\ \frac{1}{k\left(k+1\right)} = \sum_{k = 1} ^ {2n} \frac{1}{k(k+1)} - \sum_{k = 1} ^ {n-1} \frac{1}{k(k+1)} = \left(1 - \frac{1}{2n+1}\right) - \left(1 - \frac{1}{n}\right) = \frac{1}{n} - \frac{1}{2n+1} \rightarrow 0 $$

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First, let's talk about summation. The numerator can be written as (k+1)-k. Now simplifying we get [(1/k)-(1/k+1). Now applying the limits of summation we get limn→+∞ [(1/n)-(1/2n+1)] which is zero. Coming to second question it is false because you are completely messing up the question. Sum of limits can't be the limit of sum

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Notice:

  • $$\sum_{n=a}^{ac}\frac{b}{n(n+d)}=\frac{b\left[\text{H}_{ac}+\text{H}_{a+d-1}-\text{H}_{ac+d}-\text{H}_{a-1}\right]}{d}$$

So, Solving your question:

$$\sum_{n=a}^{2a}\frac{1}{n(n+1)}=\text{H}_{2a}+\text{H}_{a}-\text{H}_{2a+1}-\text{H}_{a-1}=\frac{a+1}{2a^2+a}$$

So:

$$\lim_{a\to\infty}\left[\sum_{n=a}^{2a}\frac{1}{n(n+1)}\right]=\lim_{a\to\infty}\left[\frac{a+1}{2a^2+a}\right]\lim_{a\to\infty}\left[\frac{\frac{1}{a}+\frac{1}{a^2}}{2+\frac{1}{a}}\right]=\frac{0+0}{2+0}=\frac{0}{2}=0$$

With $\text{H}_n$ is the $n^{\text{th}}$ harmonic number

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