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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, 2018 at 16:46
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    $\begingroup$ math.meta.stackexchange.com/questions/11696/… $\endgroup$
    – Rick
    Jan 9, 2018 at 16:47
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    $\begingroup$ Read the other points in Rick's link, too. $\endgroup$
    – robjohn ♦
    Jan 9, 2018 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, 2018 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, 2018 at 17:47

3 Answers 3

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Good start!

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

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

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  • $\begingroup$ Got it thanks Sir! $\endgroup$
    – Mohan Sharma
    Jan 9, 2018 at 16:38
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$\displaystyle \begin{align} \sum_{k=1}^\infty\left[\frac{1}{k!} - \frac{1}{(k+3)!}\right] &= \left\{\begin{array}{c} \dfrac{1}{1!} &+\dfrac{1}{2!} &+\dfrac{1}{3!} &+\dfrac{1}{4!} &+\dfrac{1}{5!} &+\dfrac{1}{6!} &+\dfrac{1}{7!} &+\dfrac{1}{8!} &+\dfrac{1}{9!} &+\cdots \\ & & &-\dfrac{1}{4!} &-\dfrac{1}{5!} &-\dfrac{1}{6!} &-\dfrac{1}{7!} &-\dfrac{1}{8!} &-\dfrac{1}{9!} &-\cdots \\ \end{array} \right\}\\ &=\frac{1}{1!} +\frac{1}{2!} +\frac{1}{3!} \end{align}$

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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 Alexis Gregory
    Jan 17, 2018 at 13:36

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