Using the definition:

Obviously

Mary Reyes
2021-12-22
Answered

I have to calculate the series

$\sum _{k=2}^{\mathrm{\infty}}(\frac{1}{k}-\frac{1}{k+2})$

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$\underset{n\to \mathrm{\infty}}{lim}(\frac{1}{n}-\frac{1}{n+2})=0$ , but I don't think that this is the right way to calculate the value of the series.

Using the definition:

Obviously

You can still ask an expert for help

jean2098

Answered 2021-12-23
Author has **38** answers

Let us use $\frac{1}{k}={\int}_{0}^{1}{t}^{k-1}dt$ . Then

$S=\sum _{k=2}^{\mathrm{\infty}}(\frac{1}{k}-\frac{1}{k+2})={\int}_{0}^{1}\sum _{k=2}^{\mathrm{\infty}}[{t}^{k-1}-{t}^{k+2}]dt$

$={\int}_{0}^{1}\frac{t-{t}^{3}}{1-t}dt$

$={\int}_{0}^{1}(t+{t}^{2})dt$

$=\frac{1}{2}+\frac{1}{3}$

$=\frac{5}{6}$

Foreckije

Answered 2021-12-24
Author has **32** answers

As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:

$\sum _{2}^{\mathrm{\infty}}(\frac{1}{k}-\frac{1}{k+2})=(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{6}-\frac{1}{8})+\dots$

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$\frac{5}{6}$ .

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

Therefore, the answer is

user_27qwe

Answered 2021-12-30
Author has **208** answers

Another, Different way that uses limit like you wanted to is to find the partial sum! You first compute,

Now,

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