infinite series

[logistic_guy]

Show that \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n + 1)} = 1\).

 

[khansaheb]

Show that \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n + 1)} = 1\).

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem

 

[logistic_guy]

Apply partial fraction decomposition.

\(\displaystyle \frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}\)


\(\displaystyle 1 = A(n + 1) + Bn\)

\(\displaystyle A = 1\)
\(\displaystyle B = -1\)

Then,

\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)\)

 

[pka]

Apply partial fraction decomposition.

\(\displaystyle \frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}\)


\(\displaystyle 1 = A(n + 1) + Bn\)

\(\displaystyle A = 1\)
\(\displaystyle B = -1\)

Then,

\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)\)

Can you use induction to show that if [imath]k\in Z^+[/imath] then [imath]\sum_{n=1}^{k}\frac{1}{n(n+1)}=1-\frac{1}{k+1}?[/imath]
If you can then you are done! WHY?

 

[logistic_guy]

Can you use induction to show that if [imath]k\in Z^+[/imath] then [imath]\sum_{n=1}^{k}\frac{1}{n(n+1)}=1-\frac{1}{k+1}?[/imath]
If you can then you are done! WHY?

Let me try my method first. Then, we will think about your idea!

\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right) = \lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left(\frac{1}{n} - \frac{1}{n + 1}\right)\)


\(\displaystyle = \lim_{N\rightarrow \infty}\bigg[\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N + 1}\right)\bigg]\)


\(\displaystyle = \lim_{N\rightarrow \infty}\bigg[1 - \frac{1}{N + 1}\bigg] = \textcolor{blue}{1}\)

 

[logistic_guy]

Can you use induction to show that if [imath]k\in Z^+[/imath] then [imath]\sum_{n=1}^{k}\frac{1}{n(n+1)}=1-\frac{1}{k+1}?[/imath]
If you can then you are done! WHY?

We want to prove:

\(\displaystyle \sum_{n=1}^{k}\frac{1}{n(n+1)} = 1 - \frac{1}{k + 1}\)

Base case: \(\displaystyle k = 1\).

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

And \(\displaystyle 1 - \frac{1}{2} = \frac{1}{2}\)

The base case holds. Now the inductive step. We assume this is true for \(\displaystyle k = m\).

\(\displaystyle \sum_{n=1}^{m}\frac{1}{n(n+1)} = 1 - \frac{1}{m + 1}\)

We need to show:

\(\displaystyle \sum_{n=1}^{m+1}\frac{1}{n(n+1)} = 1 - \frac{1}{m + 2}\)

\(\displaystyle \sum_{n=1}^{m+1}\frac{1}{n(n+1)} = \sum_{n=1}^{m}\frac{1}{n(n+1)} + \sum_{n=m+1}^{m+1}\frac{1}{n(n+1)} = \left(\sum_{n=1}^{m}\frac{1}{n(n+1)}\right) + \frac{1}{(m + 1)(m + 2)}\)

Using the inductive step.

\(\displaystyle \sum_{n=1}^{m+1}\frac{1}{n(n+1)} = \left(1 - \frac{1}{m + 1}\right) + \frac{1}{(m + 1)(m + 2)}\)


\(\displaystyle = 1 - \frac{m + 2}{(m + 1)(m + 2)} + \frac{1}{(m + 1)(m + 2)}\)


\(\displaystyle = 1 + \frac{-m - 2 + 1}{(m + 1)(m + 2)} = 1 + \frac{-m - 1}{(m + 1)(m + 2)}\)


\(\displaystyle = 1 - \frac{m + 1}{(m + 1)(m + 2)} = \textcolor{blue}{1 - \frac{1}{m + 2}}\)

\(\displaystyle \textcolor{red}{\bold{induction \ complete.}}\)

We have:

\(\displaystyle \sum_{n=1}^{k}\frac{1}{n(n+1)} = 1 - \frac{1}{k + 1}\)

Then,

\(\displaystyle \lim_{k\rightarrow \infty}\sum_{n=1}^{k}\frac{1}{n(n+1)} = \lim_{k\rightarrow \infty}\left(1 - \frac{1}{k + 1}\right) = \textcolor{indigo}{\bold{1}}\)