# I have to calculate the series \sum_{k=2}^\infty(\frac{1}{k}-\frac{1}{k+2}) Using the definition: L=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^n a^k Obviously \lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n+2})=0,

Mary Reyes 2021-12-22 Answered
I have to calculate the series
$\sum _{k=2}^{\mathrm{\infty }}\left(\frac{1}{k}-\frac{1}{k+2}\right)$
Using the definition:
$L=\underset{n\to \mathrm{\infty }}{lim}{S}_{n}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^{n}{a}^{k}$
Obviously $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}-\frac{1}{n+2}\right)=0$, but I don't think that this is the right way to calculate the value of the series.
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jean2098
Let us use $\frac{1}{k}={\int }_{0}^{1}{t}^{k-1}dt$. Then
$S=\sum _{k=2}^{\mathrm{\infty }}\left(\frac{1}{k}-\frac{1}{k+2}\right)={\int }_{0}^{1}\sum _{k=2}^{\mathrm{\infty }}\left[{t}^{k-1}-{t}^{k+2}\right]dt$
$={\int }_{0}^{1}\frac{t-{t}^{3}}{1-t}dt$
$={\int }_{0}^{1}\left(t+{t}^{2}\right)dt$
$=\frac{1}{2}+\frac{1}{3}$
$=\frac{5}{6}$
###### Not exactly what you’re looking for?
Foreckije
As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
$\sum _{2}^{\mathrm{\infty }}\left(\frac{1}{k}-\frac{1}{k+2}\right)=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+\dots$
From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the $\frac{1}{2}$ and $\frac{1}{3}$.
Therefore, the answer is $\frac{5}{6}$.
###### Not exactly what you’re looking for?
user_27qwe

Another, Different way that uses limit like you wanted to is to find the partial sum! You first compute, and so on... What you will find (and you can prove this fact by induction) is that
${S}_{n}=\frac{5{n}^{2}+3n-8}{6\left(n+1\right)\left(n+2\right)}$
Now,
$\underset{n\to \mathrm{\infty }}{lim}\frac{5{n}^{2}+3n-8}{6{n}^{2}+18n+12}=\frac{5}{6}$