The Mighty Taylor Series of ee.

Everybody knows the following infinite series :

∑n=1∞1n!=e−1\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!} = e-1

which comes out from the Taylor Series of exe^x. It is a single summation.

What about the following :

∑n=1∞∑m=1∞1(n+m)!\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)!}

Or, Let's generalize that for nn summations :

∑a0,a1,a2,…,an−1>01(a0+a1+a2+⋯+an−1)!\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!}

Does this have a closed form? It is a challenge for you guys!

Challenge  1  :(Easy)\mathbf{Challenge \; 1\; :} {\color{#20A900}{\textrm{(Easy)}}} : Find the value of :

∑n=1∞∑m=1∞1(n+m)!\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)!}

Challenge  2  :(Hard)\mathbf{Challenge \; 2\; :} {\color{#D61F06}{\textrm{(Hard)}}} : Find the closed form of :

∑a0,a1,a2,…,an−1>01(a0+a1+a2+⋯+an−1)!\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!}

Try them ! I will share my approach the next week.

This note is a part of the set Sequences and Series Challenges.

#Calculus #Generalize #InfiniteSequence #InfiniteSeries #Challenge

Note by Hasan Kassim
6 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Nice question.

Here's my approach:

Denote the required sum by S(n)S(n).

Focus on the denominator a0+a1+a2...an−1a_{0}+a_{1}+a_{2}...a_{n-1}.

Fix it's value to be some constant kk (for now). Consider all the terms in our summation with this denominator. How many such terms are there?

That's easy: the number of such terms is simply the number of solutions to the equation a0+a1+a2...an−1=ka_{0}+a_{1}+a_{2}...a_{n-1}=k

Note that we need solutions in the natural numbers only as all the aia_{i} vary only from 1 to infinity.

So, for a given k, the number of such terms is given by (k−1n−1)\binom{k-1}{n-1}.

Note also, that the smallest possible value of kk is n and it goes on up until infinity.

Thus, our summation can be written as:

S(n)=∑k=n∞(k−1n−1)k!S(n)=\sum_{k=n}^{\infty}\frac{\binom{k-1}{n-1}}{k!}

After simplification, this can be written as: S(n)=1(n−1)!∑k=n∞1k(k−n)!S(n)=\frac{1}{(n-1)!}\sum_{k=n}^{\infty}\frac{1}{k(k-n)!}

For convenience, let k−n=rk-n=r.

Thus, S(n)=1(n−1)!∑r=0∞1(r+n)(r)!S(n)=\frac{1}{(n-1)!}\sum_{r=0}^{\infty}\frac{1}{(r+n)(r)!}

We'll deal with the (n−1)!(n-1)! later.

For now, let T(n)=∑r=0∞1(r+n)(r)!\quad \quad T(n)=\sum_{r=0}^{\infty}\frac{1}{(r+n)(r)!}

Also, ex=∑r=0∞xrr!\quad \quad \quad e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}.

⇒xn−1ex=∑r=0∞xr+n−1r!\Rightarrow x^{n-1}e^x=\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}

⇒∫01xn−1exdx=∫01∑r=0∞xr+n−1r!dx\Rightarrow \int_{0}^{1}x^{n-1}e^xdx=\int_{0}^{1}\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}dx

Note that the RHS is the expression that we had obtained earlier for T(n)T(n).

Thus,

T(n)=∫01xn−1exdxT(n)=\int_{0}^{1}x^{n-1}e^xdx

Integrate this by parts to establish the recurrence relation: T(n)=e−(n−1)T(n−1)T(n)=e-(n-1)T(n-1)

This is a recurrence relation. Solve it and you're done. Once you've found T(n)T(n), divide that by (n−1)!(n-1)! to get S(n)S(n).

Shashwat Shukla - 6 years, 3 months ago

Log in to reply

Very elegant solution!

Just a thing: the expression you obtained from the integral for T(n)T(n) should be for T(n+1)T(n+1) instead.

Nice work :)

Hasan Kassim - 6 years, 3 months ago

Log in to reply

Thanks :)

Really enjoyed doing this problem....And yea, it should be T(n+1)T(n+1).

I've edited it now.

Shashwat Shukla - 6 years, 3 months ago

Log in to reply

@Shashwat Shukla – Also replace the coefficient of T(n−1)T(n-1) by n−1n-1, in the recursion.

Hasan Kassim - 6 years, 3 months ago

Log in to reply

@Hasan Kassim – My bad. Done now...Oh and btw, I saw this post when I was going through the problems of your set, 'Sequences and series challenges' and I liked all of your problems. Thanks for posting them :) Cheers

Shashwat Shukla - 6 years, 3 months ago

Log in to reply

@Shashwat Shukla – Yea I will keep posting new ones, my pleasure :)

Hasan Kassim - 6 years, 3 months ago

Its quite tough.

Ayush Kumar - 3 years, 1 month ago
×

Problem Loading...

Note Loading...

Set Loading...