Prove Using Induction: $\sum_{k=1}^{n} 1/k(k+1) = n/(n+1)$

Question

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)?

Shouldn't you be taking a summation of terms on the left-hand side? — paw88789, Apr 14 '16 at 17:16
It is not true that $1/k(k+1)=n/n+1$, unless you forgot something there. — Thomas Andrews, Apr 14 '16 at 17:18
Counter example $1/27(27+1) \ne 32/(32+1)$. What you wrote doesn't make any sense as k and n can each be anything. And if you restrict k = n it's obviously false. — fleablood, Apr 14 '16 at 17:27
The base case is fine, the induction step is still very badly messed up (as written in the question) because you list only one term of the necessary sum instead of showing the entire sum, and you have mixed up the use of $k$ as an induction variable with the use of $k$ as an index of the summation. But the accepted answer is good. — David K, Apr 14 '16 at 17:35

score 1 · Accepted Answer · answered Apr 14 '16 at 17:22

1

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

answered Apr 14 '16 at 17:22

openspace

6,470

yes 1/k(k+1) = n/n+1 , why do u have 1/k - 1/(k+1) = n/n+1 ? – Paul Apr 14 '16 at 17:24
@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 : $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ – openspace Apr 14 '16 at 17:27
is the induction step complete? is the question done here? – Paul Apr 14 '16 at 17:30
@Clabbe actually yes, it's better to write the base of induction but it's obviously and you could check it by yourself – openspace Apr 14 '16 at 17:31
yes! thank you! – Paul Apr 14 '16 at 17:34
@Clabbe no problem, my friend – openspace Apr 14 '16 at 17:35
sorry to bother you once again but which is the Inductive Hypothesis here? – Paul Apr 14 '16 at 19:39
@Paul suppose that $\sum{\frac{1}{k(k+1)}} = \frac{n}{n+1}$ – openspace Apr 14 '16 at 19:41

Prove Using Induction: $\sum_{k=1}^{n} 1/k(k+1) = n/(n+1)$

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