\[S = \frac{1^2}{1!} + \frac{1^2 + 2^2}{2!} + ... \\
= \sum_{n=1}^\infty[\frac{\sum_{k=1}^n k^2}{n!}] \\
= \sum_{n=1}^\infty\frac{(\frac{n(n+1)(2n+1)}{6})}{n!} \\
= \frac{1}{6}[\sum_{n=1}^\infty\frac{n(n+1)(2n+1)}{n!}] \\
= \frac{1}{6}[\sum_{n=1}^\infty\frac{(n+1)(2n+1)}{(n-1)!}] \\
= \frac{1}{6}[\sum_{n=0}^\infty\frac{(n+2)(2n+3)}{n!}]
\]
Now define \[f(x) = \sum_{n=0}^\infty [\frac{(n+2)(2n+3)}{n!}x^n].\]
\(f(x)\) is the Taylor series of \(g(x) = e^x(2x^2 + 9x + 6)\) centered at \(x=0.\) Let \(x=1,\) then \(f(1) = g(1) = e(2 + 9 + 6) = 17e.\) Also,
\[S = \sum_{n=0}^\infty\frac{(n+2)(2n+3)}{n!}\times 1^n\]
which is equal to \(f(1)\) from the definition of \(f(x).\) Therefore
\[S = \frac{1}{6}f(1) = \frac{17e}{6}\]

Edit: Wow, this note and solution have become quite popular within two hours of posting. For new viewers who are curious about any step in the proof, please see the discussion below, and if it is not explained, please comment in the discussion and I will try to answer.

I apologize if this proof is not concise, as this is the first calculus based proof of a nontrivial (or is it?) series I have ever written. I will clarify in the comments any step that anyone is unsure about. Before then, here are two things I used in the proof, but did not explicitly prove for convenience, as they are proven easily.
\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \text{ Sum of squares formula; proven by induction} \\
e^x(2x^2 + 9x + 6) = g(x) = \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}x^n. \text{ Maclaurin/Taylor series, } x_0=0
\]

Nice approach, Caleb. I had reached a similar \(f(x)\) as you but got stuck trying to establish the corresponding Taylor series. I tried taking the second derivative of the Taylor series for \(x^{2}e^{x}\) and then multiplying by \(x\) but this didn't give me the \(2n + 3\) factor. How did you determine the \((2x^{2} + 9x + 6)\) term?

@Brian Charlesworth
–
I am not 100% sure how to answer, do you mean how did I know to use \(2x^2 + 9x + 6,\) or do you mean how did I get the series expansion?

@Caleb Townsend
–
Yes, I was wondering what your method was to determine the \(2x^{2} + 9x + 6\) term. I knew that I would have to manipulate the series for \(e^{x}\) and its derivatives somehow, but wasn't able to complete the task.

@Brian Charlesworth
–
I think you were on the same track as me, I had been working for over 20 minutes on this problem so I had time to have fun. To my knowledge the easiest way to solve problems like this (polynomial in numerator, factorial in denominator) is to find a taylor series of the same degree as the polynomial in the numerator, using an exponential (usually \(e^x\)), then let \(x = 1\) so that \(x^n = 1\ \forall\ n.\) I started with \(e^x(ax^2 + bx + c)\) and got the first three terms of the Maclaurin series to be \(c + (b+c)x + (a + b + c/2)x^2.\) Comparing this to the first few values of \(\frac{(n+3)(2n+3)}{n!}x^n,\) (namely \(6,\ 15x,\ 14x^2\))
\[c = 6 \\
b+c = 15 \Rightarrow b = 9 \\
a + b + \frac{c}{2} = 14 \Rightarrow a = 2\]
I think this method is only valid for Maclaurin series as I tried it on a series about infinity and it did not yield the correct answer, but I have never retried on any other Taylor series at infinity. If so then series that don't converge at \(0\) may require different reasoning, which I have no clue of. In the words of my Calc II teacher, "math is a form of fine art."

By the way, I saw in another problem that someone asked if you are a teacher, but I never saw the answer. May I ask now if you teach mathematics? You are efficient in the way you solve problems, similar to how some professors and teachers use the fastest method that is on par with the course level, while still retaining the solution's ability to teach.

@Caleb Townsend
–
Ah, o.k., that works well. So you assume the general form and then solve for the specifics; I'll try this method now on Pi Han Goh's new question. Thanks for the explanation. :)

Math is just a recreation for me, (although I do have a background in it), and I've never taught the subject. I do try, though, to "think like a professor" when tailoring my solutions to the respective level of the questions, so I'm glad to hear that this mindset yields the desired effect. Is it a goal of yours to be a professor someday?

@Caleb Townsend
–
Oh, o.k., I never once looked at a professor and thought, "Gosh, I want to be up there teaching someday just like you!" Researching and solving real-world problems at some progressive corporation always seemed like a more attractive option if I had pursued a career in mathematics.

Well I am familiar with that problem but I do not believe the proof is concise, but the result will be the product of a Bell number and \(e.\) In fact it is \[\sum_{r=0}^\infty \frac{r^n}{r!} = B_n \times e\]
if I'm not mistaken.

Actually I would like to amend this comment, I think what you mean is
\[\sum_{r=1}^\infty \frac{\sum_{k=1}^r k^n}{r!}\]
which is not (at least in general) equal to \(B_n\times e.\) I do not know the value of that expression but Pi Han Goh has posted the case n=7, which may help generalize. His solution in particular covers use of the Bell numbers.

@Pi Han Goh
–
Neat problem, I have started to solve it but my method seems like it will take a while. I do think there may be a problem in the statement though; do you perhaps mean \(\frac{24}{e}\times W?\) The sum's numerator should be a multiple of \(e.\)

Easy Math Editor

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## Comments

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TopNewest\[S = \frac{1^2}{1!} + \frac{1^2 + 2^2}{2!} + ... \\ = \sum_{n=1}^\infty[\frac{\sum_{k=1}^n k^2}{n!}] \\ = \sum_{n=1}^\infty\frac{(\frac{n(n+1)(2n+1)}{6})}{n!} \\ = \frac{1}{6}[\sum_{n=1}^\infty\frac{n(n+1)(2n+1)}{n!}] \\ = \frac{1}{6}[\sum_{n=1}^\infty\frac{(n+1)(2n+1)}{(n-1)!}] \\ = \frac{1}{6}[\sum_{n=0}^\infty\frac{(n+2)(2n+3)}{n!}] \] Now define \[f(x) = \sum_{n=0}^\infty [\frac{(n+2)(2n+3)}{n!}x^n].\] \(f(x)\) is the Taylor series of \(g(x) = e^x(2x^2 + 9x + 6)\) centered at \(x=0.\) Let \(x=1,\) then \(f(1) = g(1) = e(2 + 9 + 6) = 17e.\) Also, \[S = \sum_{n=0}^\infty\frac{(n+2)(2n+3)}{n!}\times 1^n\] which is equal to \(f(1)\) from the definition of \(f(x).\) Therefore \[S = \frac{1}{6}f(1) = \frac{17e}{6}\]

Edit: Wow, this note and solution have become quite popular within two hours of posting. For new viewers who are curious about any step in the proof, please see the discussion below, and if it is not explained, please comment in the discussion and I will try to answer.

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I apologize if this proof is not concise, as this is the first calculus based proof of a nontrivial (or is it?) series I have ever written. I will clarify in the comments any step that anyone is unsure about. Before then, here are two things I used in the proof, but did not explicitly prove for convenience, as they are proven easily. \[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \text{ Sum of squares formula; proven by induction} \\ e^x(2x^2 + 9x + 6) = g(x) = \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}x^n. \text{ Maclaurin/Taylor series, } x_0=0 \]

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Nice approach, Caleb. I had reached a similar \(f(x)\) as you but got stuck trying to establish the corresponding Taylor series. I tried taking the second derivative of the Taylor series for \(x^{2}e^{x}\) and then multiplying by \(x\) but this didn't give me the \(2n + 3\) factor. How did you determine the \((2x^{2} + 9x + 6)\) term?

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By the way, I saw in another problem that someone asked if you are a teacher, but I never saw the answer. May I ask now if you teach mathematics? You are efficient in the way you solve problems, similar to how some professors and teachers use the fastest method that is on par with the course level, while still retaining the solution's ability to teach.

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Math is just a recreation for me, (although I do have a background in it), and I've never taught the subject. I do try, though, to "think like a professor" when tailoring my solutions to the respective level of the questions, so I'm glad to hear that this mindset yields the desired effect. Is it a goal of yours to be a professor someday?

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I knew you will come up with an interesting idea .... \(\ddot\smile\)

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indeed very nice, good, excellent, can u generalise for this \(\large{\displaystyle \sum_{r=0}^{\infty} \frac{r^n}{r!}}\).

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Well I am familiar with that problem but I do not believe the proof is concise, but the result will be the product of a Bell number and \(e.\) In fact it is \[\sum_{r=0}^\infty \frac{r^n}{r!} = B_n \times e\] if I'm not mistaken.

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You are not mistaken

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Actually I would like to amend this comment, I think what you mean is \[\sum_{r=1}^\infty \frac{\sum_{k=1}^r k^n}{r!}\] which is not (at least in general) equal to \(B_n\times e.\) I do not know the value of that expression but Pi Han Goh has posted the case n=7, which may help generalize. His solution in particular covers use of the Bell numbers.

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Can you elaborate on how you get \(2x^2 +9x+6 \)?

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I have now posted the method under Brian Charlesworth's comment in this discussion.

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My problem inspired by you!

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@Brian Charlesworth

Yep, typo, thanks!Log in to reply

@Calvin Lin sir, plz help by posting a solution

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I don't really understand your expression, but isn't anything plus \(\infty\) equals \(\infty\)?

You should change your expression into a summation :)

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