I got to this question which is extremely hard:
Calculate:
$$\frac{1}{1*3}+\frac{1}{2*4}+\frac{1}{3*5}+...$$
How would you solve a problem like this.
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Hint: $\frac{1}{1 \times 3} = \frac{1}{2} \left(\frac{1}{1} - \frac{1}{3}\right)$ – Xiangxiang Xu Sep 28 '18 at 07:01
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Related Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ – Sil Sep 28 '18 at 07:18
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Hm I cannot vote for another duplicate after canceling previous, interesting. Anyway this looks like a duplicate Show convergence of a given series and find the limit. – Sil Sep 28 '18 at 07:22
3 Answers
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Hint: let $s_n:= \sum_{k=1}^n \frac{1}{k(k+2)}$ and observe that $\frac{1}{k(k+2)}=\frac{1}{2}(\frac{1}{k}-\frac{1}{k+2})$.
Can you proceed ?
Fred
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@BiliDebili No. This is a telescoping sum. Try writing out the first few partial sums to see what happens. – N. F. Taussig Sep 28 '18 at 07:09
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let is consider your sum $$S=\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot 7}+\dots$$ Multiply $S$ by two: $$2S=\frac{2}{1\cdot 3}+\frac{2}{2\cdot 4}+\frac{2}{3\cdot 5}+\frac{2}{4\cdot6}+\frac{2}{5\cdot 7}+\dots$$ Per fraction, write each numerator as the difference between the two components in the corresponding denominator, i.e. $2=3-1,2=4-2,2=5-3,2=6-4,2=7-5,\dots$: $$2S=\frac{3-1}{1\cdot 3}+\frac{4-2}{2\cdot 4}+\frac{5-3}{3\cdot 5}+\frac{6-4}{4\cdot6}+\frac{7-5}{5\cdot 7}+\dots$$ Split each fraction and simplify the expression: $$\begin{align*} 2S&=\left[\frac{3}{1\cdot3}-\frac{1}{1\cdot3}\right]+\left[\frac{4}{2\cdot4}-\frac{2}{2\cdot4}\right]+\left[\frac{5}{3\cdot5}-\frac{3}{3\cdot5}\right]+\left[\frac{6}{4\cdot6}-\frac{4}{4\cdot6}\right]+\left[\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right]+\dots \\ &=\left[\frac{1}{1}\color{red}{-\frac{1}{3}}\right]+\left[\frac{1}{2}\color{red}{-\frac{1}{4}}\right]\color{red}{+\left[\frac{1}{3}-\frac{1}{5}\right]+\left[\frac{1}{4}-\frac{1}{6}\right]+\left[\frac{1}{5}-\frac{1}{7}\right]+\dots} \\ &=\frac{1}{1}+\frac{1}{2} \\ &=\frac{3}{2} \\ \end{align*}$$ Divide by two to obtain a result for $S$: $$S=\frac{3}{4}$$ Thus, $$\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot 7}+\dots=\frac{3}{4}$$
Jameson
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The partial sums of $\sum_{k=1}^{\infty}\frac{1}{k\left(k+2\right)}$ can be expressed as:
$$\sum_{k=1}^{n} \frac{1}{k\left(k+2\right)} = \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$
The limit can be obtained:
$$\frac{3}{4}-\lim_{n\to\infty}\dfrac{2n+3}{2(n+1)(n+2)}$$ And then:
$$\lim_{n\to\infty}\dfrac{2n+3}{2(n+1)(n+2)} = \dfrac{1}{2} \lim_{n\to\infty}\dfrac{2n+3}{(n+1)(n+2)} \\ = \dfrac{\lim_{n\to\infty}\dfrac{2n+3}{n^2+3n+2}}{2}=\lim_{n\to\infty}\dfrac{\dfrac{\dfrac{2}{n}+\dfrac{3}{n^2}}{1+\dfrac{3}{n}+\dfrac{2}{n^2}}}{2}$$
And the limits of $\frac{2}{n},\frac{2}{n^2},\frac{3}{n}$ all tend to zero as $n\to\infty$. So then we have
$$\dfrac{3}{4} - \dfrac{0}{2}$$
Joseph Eck
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