Series and Integral
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Series and Integral
Calculate :
1. \(\displaystyle{\int \dfrac{2\,x+3}{x\,(x+1)\,(x+2)\,(x+3)+2015}\,\mathrm{d}x}\)
2. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}}\)
1. \(\displaystyle{\int \dfrac{2\,x+3}{x\,(x+1)\,(x+2)\,(x+3)+2015}\,\mathrm{d}x}\)
2. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}}\)
- Tolaso J Kos
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Re: Series and Integral
We see that the series telescopes since:Papapetros Vaggelis wrote: 2. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}}\)
$$\sum_{n=1}^{\infty}\frac{n}{(n+1)!}= \sum_{n=1}^{\infty}\left ( \frac{1}{n!}- \frac{1}{(n+1)!} \right )= 1- \lim_{n \rightarrow +\infty} \frac{1}{(n+1)!}= 1$$
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- Tolaso J Kos
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Re: Series and Integral
\begin{align*}Papapetros Vaggelis wrote:Calculate :
1. \(\displaystyle{\int \dfrac{2\,x+3}{x\,(x+1)\,(x+2)\,(x+3)+2015}\,\mathrm{d}x}\)
\int \frac{2x+3}{x(x+1)(x+2)(x+3)+2015} \, {\rm d}x&=\int \frac{2x+3}{\left ( x^2+3x \right )\left ( x^2+3x+2 \right )+2015}\, {\rm d}x \\
&\overset{u=x^2+3x}{=\! =\! =\! =\! =\! }\int \frac{1}{u(u+2)+2015}\, {\rm d}u \\
&= \int\frac{1}{2014 + \left ( u+1 \right )^2}\, {\rm d}u\\
&= \frac{1}{2014}\int \frac{{\rm d}u}{1+ \left ( \frac{u+1}{\sqrt{2014}} \right )^2}\\
&=\cdots \\
&= \frac{1}{\sqrt{2014}}\tan^{-1}\left ( \frac{x^2+3x+1}{\sqrt{2014}} \right )+c, \;\; c \in \mathbb{R}
\end{align*}
Imagination is much more important than knowledge.
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Re: Series and Integral
2. Here is another solution.
Firstly, \(\displaystyle{0<a_{n}=\dfrac{n}{(n+1)!}\leq \dfrac{n!}{(n+1)!}=\dfrac{1}{n+1}\,,\forall\,n\in\mathbb{N}}\) .
Since, \(\displaystyle{\dfrac{1}{n+1}\longrightarrow 0\,,n\longrightarrow +\infty}\), we get : \(\displaystyle{\dfrac{n}{(n+1)!}\longrightarrow 0\,,n\longrightarrow +\infty}\) .
Also, the series converges because :
\(\displaystyle{\begin{aligned} \dfrac{a_{n+1}}{a_{n}}&=\dfrac{\displaystyle{\dfrac{n+1}{(n+2)!}}}{\displaystyle{\dfrac{n}{(n+1)!}}}\\&=\left(1+\dfrac{1}{n}\right)\,\dfrac{1}{n+2}\longrightarrow 0\end{aligned}}\)
Now,
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=\sum_{n=1}^{\infty}\left(\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\right)}\)
where
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{n!}=\sum_{n=0}^{\infty}\dfrac{1^{n}}{n!}-1=e-1\,\,(I)}\)
and
\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\\&=\sum_{n=1}^{\infty}\dfrac{1}{n!}-1\\&\stackrel{(I)}{=}e-2\end{aligned}}\)
Finally,
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=\sum_{n=1}^{\infty}\dfrac{1}{n!}-\sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}=e-1-e+2=1}\)
Firstly, \(\displaystyle{0<a_{n}=\dfrac{n}{(n+1)!}\leq \dfrac{n!}{(n+1)!}=\dfrac{1}{n+1}\,,\forall\,n\in\mathbb{N}}\) .
Since, \(\displaystyle{\dfrac{1}{n+1}\longrightarrow 0\,,n\longrightarrow +\infty}\), we get : \(\displaystyle{\dfrac{n}{(n+1)!}\longrightarrow 0\,,n\longrightarrow +\infty}\) .
Also, the series converges because :
\(\displaystyle{\begin{aligned} \dfrac{a_{n+1}}{a_{n}}&=\dfrac{\displaystyle{\dfrac{n+1}{(n+2)!}}}{\displaystyle{\dfrac{n}{(n+1)!}}}\\&=\left(1+\dfrac{1}{n}\right)\,\dfrac{1}{n+2}\longrightarrow 0\end{aligned}}\)
Now,
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=\sum_{n=1}^{\infty}\left(\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\right)}\)
where
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{n!}=\sum_{n=0}^{\infty}\dfrac{1^{n}}{n!}-1=e-1\,\,(I)}\)
and
\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\\&=\sum_{n=1}^{\infty}\dfrac{1}{n!}-1\\&\stackrel{(I)}{=}e-2\end{aligned}}\)
Finally,
\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=\sum_{n=1}^{\infty}\dfrac{1}{n!}-\sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}=e-1-e+2=1}\)
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