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I wanted to find the sum $\displaystyle\sum_{k=1}^n\frac{1}{k^2+k}.$ I have proven using induction that the sum is $\frac{n}{n+1}$, but I was wondering if there was another method that didn't use induction.

AreS
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Yes, there is a very elegant way to evaluate the sum without using induction. This sum is a classic telescoping sum.

First, notice that $\frac{1}{k^2+k}=\frac{1}{k}-\frac{1}{k+1}.$ Now, we label the sum $S_n$ and rewrite it using the observation:

$$S_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}.$$

Notice that every term other than the first and the last get cancelled out. So,

$$S_n=1-\frac{1}{n+1}=\frac{n}{n+1}.$$

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    Of course, the correct way to prove a telescoping sum is ... using induction. – Thomas Andrews Dec 31 '22 at 17:20
  • @ThomasAndrews, I'm not sure I follow. What do you mean by "prove a telescoping sum"? Is my method not rigorous enough? –  Dec 31 '22 at 17:41
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    If $g(n)=f(n)-f(n+1)$ then $\sum_{k=1}^n g(k)=f(1)-f(n+1).$ We can show it intuitive by writing out the sum and showing the cancellation, but a formal proof uses induction. Indeed, $\sum_{k=1}^n$ is inductively defined, on $n,$ so you require induction to prove practically anything involving sums. – Thomas Andrews Dec 31 '22 at 17:51