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The task at hand is to prove using induction that the following proposition holds for all $n \in \mathbb{N}$.

$$P(n): \sum_{k=1}^{n}1/k(k+1) = n/(n+1)$$

Here is the proof I have thus far:

Base Case: $(n=1)$

LHS $= 1/1(1+1) = 1/(1+1) =$ RHS

i.e., both LHS and RHS are $1/2$.

So the base case is true.

Step Case: Assume the proposition holds $n=k$; then show that it holds for $n=k+1$.

$1/k(k+1) + 1/(k+1)(k+2)$

$= k/(k+1) + 1/(k+1)(k+2)$

$= k(k+2)/(k+1)(k+2) + 1/(k+1)(k+2)$

$= k(k+2)+ 1/(k+1)(k+2)$

$= k^2 + 2k + 1$

$= (k + 1)^2 / (k+1) (k+2)$

$= (k+1)/(k+2)$

$= (k+1)/((k+1)+1)$

Is this the right way to do it?

If so, where could my proof be clearer?

If not, could you help me to find the error(s)?

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    $\begingroup$ Shouldn't you be taking a summation of terms on the left-hand side? $\endgroup$ – paw88789 Apr 14 '16 at 17:16
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    $\begingroup$ It is not true that $1/k(k+1)=n/n+1$, unless you forgot something there. $\endgroup$ – Thomas Andrews Apr 14 '16 at 17:18
  • $\begingroup$ i think you forgot a sum $\endgroup$ – Dr. Sonnhard Graubner Apr 14 '16 at 17:19
  • $\begingroup$ Please write your math in $\LaTeX$. $\endgroup$ – Jacob Apr 14 '16 at 17:23
  • $\begingroup$ $\LaTeX$ help. $\endgroup$ – user 170039 Apr 14 '16 at 17:26
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I think you mean the $\sum{\frac{1}{k(k+1)}}$

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

But if you want induction:

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

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  • $\begingroup$ yes 1/k(k+1) = n/n+1 , why do u have 1/k - 1/(k+1) = n/n+1 ? $\endgroup$ – Paul Apr 14 '16 at 17:24
  • $\begingroup$ @Clabbe as you can see : $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, consider first few parts of the sum: $1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$ , so we see that some parts destoy each other, so you've got : $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ $\endgroup$ – openspace Apr 14 '16 at 17:27
  • $\begingroup$ is the induction step complete? is the question done here? $\endgroup$ – Paul Apr 14 '16 at 17:30
  • $\begingroup$ @Clabbe actually yes, it's better to write the base of induction but it's obviously and you could check it by yourself $\endgroup$ – openspace Apr 14 '16 at 17:31
  • $\begingroup$ yes! thank you! $\endgroup$ – Paul Apr 14 '16 at 17:34

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