\[S = \frac{1^2}{1!} + \frac{1^2 + 2^2}{2!} + ... \\
= \sum_{n=1}^\infty[\frac{\sum_{k=1}^n k^2}{n!}] \\
= \sum_{n=1}^\infty\frac{(\frac{n(n+1)(2n+1)}{6})}{n!} \\
= \frac{1}{6}[\sum_{n=1}^\infty\frac{n(n+1)(2n+1)}{n!}] \\
= \frac{1}{6}[\sum_{n=1}^\infty\frac{(n+1)(2n+1)}{(n-1)!}] \\
= \frac{1}{6}[\sum_{n=0}^\infty\frac{(n+2)(2n+3)}{n!}]
\]
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.
–
Caleb Townsend
·
1 year, 9 months ago

Log in to reply

@Caleb Townsend
–
I knew you will come up with an interesting idea .... \(\ddot\smile\)
–
Nihar Mahajan
·
1 year, 9 months ago

Log in to reply

@Caleb Townsend
–
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.
\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \text{ Sum of squares formula; proven by induction} \\
e^x(2x^2 + 9x + 6) = g(x) = \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}x^n. \text{ Maclaurin/Taylor series, } x_0=0
\]
–
Caleb Townsend
·
1 year, 9 months ago

Log in to reply

@Caleb Townsend
–
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
·
1 year, 9 months ago

Log in to reply

@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
·
1 year, 9 months ago

Log in to reply

@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
·
1 year, 9 months ago

Log in to reply

@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
·
1 year, 9 months ago

Log in to reply

@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?
–
Brian Charlesworth
·
1 year, 9 months ago

Log in to reply

@Brian Charlesworth
–
Haha well it is the "backup plan," in fact I couldn't realistically be a professor without getting speech therapy first.
–
Caleb Townsend
·
1 year, 9 months ago

Log in to reply

@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.
–
Brian Charlesworth
·
1 year, 9 months ago

@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 \(e.\)
–
Caleb Townsend
·
1 year, 9 months ago

@Pi Han Goh
–
Great new question, but I think that you meant to write \(\dfrac{24}{e} \times W.\) :)
–
Brian Charlesworth
·
1 year, 9 months ago

Log in to reply

@Caleb Townsend
–
indeed very nice, good, excellent, can u generalise for this \(\large{\displaystyle \sum_{r=0}^{\infty} \frac{r^n}{r!}}\).
–
Tanishq Varshney
·
1 year, 9 months ago

Log in to reply

@Tanishq Varshney
–
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.
–
Caleb Townsend
·
1 year, 9 months ago

Log in to reply

@Tanishq Varshney
–
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.
–
Caleb Townsend
·
1 year, 9 months ago

## Comments

Sort by:

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. – Caleb Townsend · 1 year, 9 months ago

Log in to reply

– Nihar Mahajan · 1 year, 9 months ago

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

– Caleb Townsend · 1 year, 9 months ago

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. \[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \text{ Sum of squares formula; proven by induction} \\ e^x(2x^2 + 9x + 6) = g(x) = \sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}x^n. \text{ Maclaurin/Taylor series, } x_0=0 \]Log in to reply

– Brian Charlesworth · 1 year, 9 months ago

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?Log in to reply

– Caleb Townsend · 1 year, 9 months ago

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?Log in to reply

– Brian Charlesworth · 1 year, 9 months ago

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.Log in to reply

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 · 1 year, 9 months ago

Log in to reply

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? – Brian Charlesworth · 1 year, 9 months ago

Log in to reply

– Caleb Townsend · 1 year, 9 months ago

Haha well it is the "backup plan," in fact I couldn't realistically be a professor without getting speech therapy first.Log in to reply

– Brian Charlesworth · 1 year, 9 months ago

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.Log in to reply

– Pi Han Goh · 1 year, 9 months ago

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

– Caleb Townsend · 1 year, 9 months ago

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

My problem inspired by you! – Pi Han Goh · 1 year, 9 months ago

Log in to reply

– Caleb Townsend · 1 year, 9 months ago

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 \(e.\)Log in to reply

@Brian Charlesworth – Pi Han Goh · 1 year, 9 months ago

Yep, typo, thanks!Log in to reply

– Brian Charlesworth · 1 year, 9 months ago

Great new question, but I think that you meant to write \(\dfrac{24}{e} \times W.\) :)Log in to reply

– Tanishq Varshney · 1 year, 9 months ago

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

Pi Han Goh has posted the case n=7, which may help generalize. His solution in particular covers use of the Bell numbers. – Caleb Townsend · 1 year, 9 months ago

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 butLog in to reply

– Caleb Townsend · 1 year, 9 months ago

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.Log in to reply

You are not mistaken – Pi Han Goh · 1 year, 9 months ago

Log in to reply

– Tanishq Varshney · 1 year, 9 months ago

thanx for the help \(\ddot \smile\)Log in to reply

@Calvin Lin sir, plz help by posting a solution – Tanishq Varshney · 1 year, 9 months ago

Log in to reply

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

You should change your expression into a summation :) – Trung Đặng Đoàn Đức · 1 year, 9 months ago

Log in to reply