Addition Addiction

Calculus Level 4

$\left \lceil \displaystyle \sum_{a=1}^{5} \left ( \displaystyle \sum_{b=1}^{\infty} \displaystyle \dfrac{b^a}{b!} \right ) \right \rceil = \: \: ?$

This question was inspired by a Harvard-MIT Mathematics Tournament Calculus question.

The answer is 204.

2 solutions

Guilherme Dela Corte
Apr 13, 2014

I'll use HMMT's solution.

Let $S_n = \displaystyle \sum_{k=0}^{\infty} \frac{k^n}{k!}$ , and for convenience, $0^0 = 1$ . We have

$S_n = \sum_{k=0}^{\infty} \frac{k^n}{k!} = \sum_{k=1}^{\infty} \frac{k^n}{k!} = \sum_{k=0}^{\infty} \frac{(k+1)^n}{(k+1)!} = \sum_{k=0}^{\infty} \frac{(k+1)^{n-1}}{k!} = \sum_{i=0}^{n-1} \binom{n-1}{i} S_i$

We can compute inductively that $S_1 = e, S_2 = 2e, S_3 = 5e, S_4 = 15e, S_5=52e$ . Summing it all, we obtain $75e \approx \boxed{204.}$

I got there much less elegantly. Basically I used that $\sum_{b=1}^{\infty} \frac{b(b-1)\cdots(b-k)}{b!} = e$ for all nonnegative integers $k$ , and if you expand out, you can get a formula for $S_{k+1}$ in terms of the lower ones.

This sequence seems to be quite well-known: http://oeis.org/A000110

Patrick Corn - 7 years, 1 month ago

Sorry, can you recommend me a good book text to study these matters? I'd appreciate it :)

Carlos David Nexans - 6 years, 10 months ago

Uttaran Choudhurry
Apr 21, 2014

S1=e, S2=2e, S3=5e, S4=15e, S5=52e adding up all gives 75e ~ 204

Uttaran, how did you find these values of $S_n$ ?

Guilherme Dela Corte - 7 years, 1 month ago

Addition Addiction

This question was inspired by a Harvard-MIT Mathematics Tournament Calculus question.

The answer is 204.

2 solutions

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