Compute $$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$

My Approach

Since $k^{3}+6k^{2}+11k+5= \left(k+1\right)\left(k+2\right)\left(k+3\right)-1$

$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!} = \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{\left(k+3\right)!}\right)$$

But now I can't find this limit.

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    $\begingroup$ It is better to use MathJax for math formulas. That is why it is there. Images are for illustrations, not formulas since we have MathJax. It is easier to search MathJax than an image. It is easier to read and edit MathJax, too. Please do not use images for formulas. $\endgroup$ – robjohn Jan 9 '18 at 16:46
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    $\begingroup$ math.meta.stackexchange.com/questions/11696/… $\endgroup$ – Rick Jan 9 '18 at 16:47
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    $\begingroup$ Read the other points in Rick's link, too. $\endgroup$ – robjohn Jan 9 '18 at 16:49
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    $\begingroup$ The main advantage of MathJax over images is that the content can be searched, so please do not rollback such improvement. $\endgroup$ – Jack D'Aurizio Jan 9 '18 at 17:14
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    $\begingroup$ A lesser, but still not insignifcant advantage of MathJax over a picture is that any answerer can copy/paste the source code of the formula to their answer. Saving their precious time for something more useful. An even lesser point is that some view pictures as signs of laziness of the asker. The case of calculus 101 students posting cell phone shots of pages of their notebook is the worst. Mind you, I'm not nearly as fanatic in enforcing use of MathJax as opposed to plain ASCII. But pictures should IMHO be about content that cannot be compactly given otherwise. Like, "worth a thousand words". $\endgroup$ – Jyrki Lahtonen Jan 9 '18 at 17:47

Good start!

$$ \begin{align} \lim_{n \to \infty}\sum_{k=1}^n\left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=1}^n\frac{1}{(k+3)!} \right) \\ &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=4}^{n+3}\frac{1}{k!} \right) \\ &= \lim_{n\to\infty}\left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - \frac{1}{(n+1)!} - \frac{1}{(n+2)!} - \frac{1}{(n+3)!} \right) \\ &= \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \\ &= \frac{5}{3} \end{align} $$


Use $$\sum_{k=1}^{\infty}\frac{1}{k!}=e-1$$ and $$\sum_{k=1}^{\infty}\frac{1}{(k+3)!}=e-2-\frac{1}{2}-\frac{1}{6}$$

  • $\begingroup$ Got it thanks Sir! $\endgroup$ – Mohan Sharma Jan 9 '18 at 16:38

$\displaystyle \begin{align} \sum_{k=1}^\infty\left[\frac{1}{k!} - \frac{1}{(k+3)!}\right] &= \left\{\begin{array}{c} \dfrac{1}{1!} &+\dfrac{1}{2!} &+\dfrac{1}{3!} &+\dfrac{1}{4!} &+\dfrac{1}{5!} &+\dfrac{1}{6!} &+\dfrac{1}{7!} &+\dfrac{1}{8!} &+\dfrac{1}{9!} &+\cdots \\ & & &-\dfrac{1}{4!} &-\dfrac{1}{5!} &-\dfrac{1}{6!} &-\dfrac{1}{7!} &-\dfrac{1}{8!} &-\dfrac{1}{9!} &-\cdots \\ \end{array} \right\}\\ &=\frac{1}{1!} +\frac{1}{2!} +\frac{1}{3!} \end{align}$

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    $\begingroup$ It should be mentioned that the series is absolutely convergent, hence its terms can be rearranged without changing its value. $\endgroup$ – steven gregory Jan 17 '18 at 13:36

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