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Let $n\geq1$. Find a formula for the sum:

$$S_n=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{n\cdot(n+1)}$$

The formula for the sum should be based on n components of the last term:

$$S_n= n\cdot\frac{1}{n(n+1)}$$

So that should give:

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

Proof by induction, consider the base case $n=1$:

$$\sum_{n=1}^1\frac{1}{(1+1)}=\frac{1}{2}$$

But how do I prove the case for $n+1$? Thanks

Luthier415Hz
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    Wrong yes, but absolutely no reason for downvotes here. – Paul Jul 13 '23 at 09:00
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    Find $S_1$, $S_2$, $S_3$, and spot the pattern to give an inductive hypothesis for $S_n$. It is not $n\cdot\frac{1}{n(n+1)}$ so you may have made a slip. Then prove it using $S_{k+1}=S_k+\frac{1}{(k+1)(k+2)}$ – Henry Jul 13 '23 at 09:05
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    The sum is not n times a general term. That only works if each term is constant, which it clearly isn't. Note instead that $\frac{1}{2.3}=\frac{1}{2} -\frac{1}{3}$. Can you repeat that for the other terms? Do you see how to then sum them? – Paul Jul 13 '23 at 09:07
  • Writing a sum of a series without asking yourself if it is convergent or divergent is nonsense. All series you are talking about are unfortunately divergent so you have to see the problem otherwise.

    Notice that for all $k>0$, $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ and try to compute the sum $S_n$.

     – Bebop Jul 13 '23 at 09:08

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