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In the problem below, I got the partial fraction decomposition but don't know how to get the answer:

$$\sum^\infty_{k=1}{\frac{1}{k(k+2)}}$$

I found the answer, using an online calculator, to be $\frac{3}{4}$.

When I did the partial fraction decomposition I got:

$$\sum^\infty_{k=1}\frac{1}{2k}-\frac{1}{2(k+2)}$$

Then I got the sum of

$$\left(\frac{1}{2}\right)+\dotsi-\frac{1}{2(n+2)}$$

Why does this equal $\frac{3}{4}$ and not $\frac{1}{2}$?

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We have that

$$\sum^\infty_{k=1}{\frac{1}{k(k+2)}}=\sum^\infty_{k=1}\frac12\left({\frac{1}{k}}-{\frac{1}{k+2}}\right)=\frac12\left(1-\frac13+\frac12-\frac14+\frac13-\frac15+\ldots\right)=\frac12\left(\frac32\right)$$

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For $n \geq 3,$ the partial sums $$\begin{align*}\sum_{k=1}^n \left(\frac{1}{2k} - \frac{1}{2(k+2)}\right) &= \sum_{k=1}^n \frac{1}{2k} - \sum_{k=1}^n \frac{1}{2(k+2)} \\ &= \sum_{k=1}^n \frac{1}{2k} - \sum_{k=3}^{n+2} \frac{1}{2k} \\ &= \left(\frac{1}{2(1)} + \frac{1}{2(2)} + \sum_{k=3}^n \frac{1}{2k}\right) - \left(\sum_{k=3}^n \frac{1}{2k} + \frac{1}{2(n+1)} + \frac{1}{2(n+2)}\right) \\ &= \frac{1}{2(1)} + \frac{1}{2(2)} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)} \\ &= \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}\end{align*}$$ have limit $\frac{3}{4}.$

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In

$$\sum^\infty_{k=1}\frac{1}{2k}-\frac{1}{2(k+2)},$$

all fractions cancel out with their negative ($\pm$) except for two. Which two?

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$\require{cancel}$The $n^{\textrm{th}}$ partial sum is $$S_n =\sum\limits_{k=1}^n\left(\frac{1}{2k}-\frac{1}{2(k+2)}\right)$$ $$=\sum\limits_{k=1}^n\frac{1}{2k} - \sum\limits_{k=1}^n\frac{1}{2(k+2)}$$ $$=\sum\limits_{k=1}^n\frac{1}{2k} - \sum\limits_{k=3}^{n+2}\frac{1}{2k}$$ $$=\left(\underbrace{\frac12 + \frac14}_{\textrm{terms for }k=1,2} + \cancel{\sum\limits_{k=3}^n\frac{1}{2k}}\right)-\left( \cancel{\sum\limits_{k=3}^{n}\frac{1}{2k}}+\underbrace{\frac1{n+1}+\frac1{n+2}}_{\textrm{terms for }k=n+1,n+2}\right)$$ $$=\frac12 + \frac14-\frac1{n+1}-\frac1{n+2}$$ $$=\frac34 -\frac1{n+1}-\frac1{n+2}$$ So the infinite series is, by definition, the limit of the partial sums: $$\sum\limits_{k=1}^{\infty}\left(\frac{1}{2k}-\frac{1}{2(k+2)}\right) \stackrel{\textrm{def}}{=} \lim\limits_{n\to\infty}S_n = \boxed{\frac34}$$ Note that finding the limit of the partial sums is the ONLY correct way to find the sum of the infinite series. An infinite series is a limit, not a sum.

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From the definition of the harmonic number:

$$H_n=\sum_{k=1}^\infty\frac{n}{k(k+n)}$$

it follows that

$$\sum_{k=1}^\infty\frac{1}{k(k+2)}=\frac{H_2}{2}=\frac{1+1/2}{2}=\frac34$$


Anoher method is by converting the sum to integral

$$\sum_{k=1}^\infty\frac{1}{k(k+2)}=\sum_{k=1}^\infty\frac1k\int_0^1x^{k+1}dx=\int_0^1x\left(\sum_{k=1}^\infty\frac{x^k}{k}\right)dx\\=-\int_0^1 x\ln(1-x)\ dx=-\int_0^1 (1-x)\ln x\ dx\\=-\int_0^1\ln x\ dx+\int_0^1 x\ln x\ dx\\=-(-1)-\frac14=\frac34$$

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In the partial fraction decomposition: $$\frac1{n(n+2)}=\frac12\biggl(\frac1n-\frac1{n+2}\biggr)$$ the first term is cancelled by the term with a minus sign in the second group before this group, and conversely the term with a minus sign is cancelled by the first term in the second group after: $$\sum_{k\ge 1}\frac1{k(k+2}=\frac12\biggr[\biggl(\frac11-\frac13\biggr)+\biggl(\frac12-\frac 14\biggr)+\biggl(\frac13-\frac15\biggr)+\biggl(\frac14-\frac16\biggr)+\dotsb\:\biggr], $$ so there remains in the end: $$\frac12\biggl(\frac11+\frac12\biggr)=\frac 34.$$

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