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Happened to stumble across this question and to me it immediately made me assume it's a proof my induction question but doesn't seem to be so.

Question : Show that for every natural $k$ and $n$ we have $$\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)}$$ Hence deduce that the sum above is smaller than $1/k$

I tried to do this by proof by induction but I happened to find no way to link the $n=k$ and $n+k+1$ part and unless I made a stupid error, I don't see how this can be proved by induction.

Does anyone know how one would answer this question? Thanks.

Robert Z
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SharpieMaster
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3 Answers3

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Hint: You can fix $k \in \mathbb{N}^*$ and show by induction on $n$ $$ \mathcal{P}_n : \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)} $$ To show $\mathcal{P}_n \implies \mathcal{P}_{n+1}$, you have to prove that $$ \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)} + \frac{1}{(n+k)(n+k+1)}= \frac{n+1}{k(n+1+k)} $$ using $\mathcal{P}_n$.

Monadologie
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$$\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\frac{1}{k(k+1)}$$ That's true for all of the terms of the sequence, so $$\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\frac{1}{(k+2)(k+3)}+...+\frac{1}{(k+n-1)(k+n)} $$ $$=\left(\frac{1}{k}-\frac{1}{k+1}\right)+\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\left(\frac{1}{k+2}-\frac{1}{k+3}\right)+...+\left(\frac{1}{k+m-1}- \frac{1}{k+n}\right)$$ $$=\frac{1}{k}-\frac{1}{k+n}=\frac{k+n}{k(k+n)}-\frac{k}{k(k+n)}=\frac{n}{k(k+n)}$$

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Let $P(n)$ defined as below be true. $$P(n): \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)}$$

Then, $$P(n+1) : LHS = \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}+\frac{1}{(n+k)(n+k+1)}$$ $$\implies LHS = \frac{n}{k(n+k)}+\frac{1}{(n+k)(n+k+1)} = \frac{n(n+k+1)+k}{k(n+k)(n+k+1)}=\frac{(n+1)(n+k)}{k(n+k)(n+k+1)}=\frac{n+1}{k(n+k+1)}=RHS$$

So, $P(n)$ is true $\implies P(n+1)$ is true

19aksh
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