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I've been trying to find the sum of the following infinite series: $$ \sum\limits_{n=1}^\infty \frac{x^n}{n!2^n} $$

I've rewritten it as $$\sum\limits_{n=1}^\infty \frac{y^n}{n!}, y=\frac{x}{2}$$ which I know from looking at a table has the solution $$ S_\infty = e^y - 1$$

Edit: I need to be able to show this without already knowing the answer

However, I don't know how to get from the summation to the solution in order to show work. I tried taking a look at this solution to a similar problem, but I couldn't find a way to properly apply the concepts to this one. I'd appreciate a push in the right direction for this problem.

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Revue
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2 Answers2

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Your approach is almost correct, assuming that you know the fact that

$$\sum\limits_{n = 0}^{\infty} \frac{y^n}{n!} = e^y$$

for every $y \in \mathbb{R}$ (or $\mathbb{C}$ for that matter); in particular, we can choose $y = x/2$ here, and that's all that needs to be said about the manipulation of the series. However, note that the index starts at $0$, whereas your problem starts at $n = 1$; to fix this, we add and subtract the $n = 0$ term as follows:

$$\sum\limits_{n = 1}^{\infty} \frac{y^n}{n!} = \sum\limits_{n = 0}^{\infty} \frac{y^n}{n!} - \frac{y^0}{0!} = e^y - 1$$

  • nicely explained – Betty Mock Jan 16 '14 at 23:46
  • thanks, I guess I didn't make it clear, but I need a method without already knowing the answer. – Revue Jan 17 '14 at 00:53
  • @Revue The easiest method would likely be to simply prove the formula for $e^y$; these formulas are rather equivalent. –  Jan 17 '14 at 00:54
  • The only method I can think of right now would be to write the maclaurin series for $e^y$ which will result in the sum, but I don't know where to start to work in the opposite direction. – Revue Jan 17 '14 at 01:32
  • @Revue What do you mean "the opposite direction"? If you can use the Maclaurin series to show that $e^y = 1 + y + y^2/2! + ...$, then you're done. –  Jan 17 '14 at 01:48
  • Where can I find the deduction that the particular sum given (with a factorial denominator) is equal to e^y? – Nermin Nov 26 '21 at 08:21
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Another fascinating derivative of this same thing is in the case of y=-1 which converges to e^-1 = 1/e,

That being 1/0! -1/1! + 1/2! -1/3! + 1/4! -1/5!....etc

That’s because this equals e^(i^2) And e^xi = cos(x) + isin(x), Euler’s equation And in the case of x = i itself

Sin(x) = x - x^3/3! + x^5/5! - x^7/7!... Etc Cos(x) = d/dx Sin(x) = 1 -x^2/2! + x^4/4! -x^6/6!...etc

So sin(i) = i - i^3/3! + i^5/5! ...etc = i -(-i)/3! +i/5! -(-i)/7! + i/9! Since i ^(n+/-4 ) always = i^n And i ^1 =i i ^2=-1 i ^3= -i i ^4=1 i ^5 = i^1 and so forth So sin(i) = i * (1 + 1/3! + 1/5! + 1/7! + 1/9!... And so i*sin(i) = -1 (1 + 1/3! + 1/5! ....)

Cos(i) = 1 -i ^2/2! + i ^4/4! - i ^6/6!

= 1 - (-1/2!) + 1/4! -(-1/6!) ....etc

1 + 1/2! + 1/4! + 1/6! + 1/8!

And so if 1/e = e^-1 = e^(i *i) = cos(i) + i *sin(i)

Then 1 -1 + 1/2! -1/3! + 1/4! -1/5! +1/6!.... = cos(i) + i sin(i) = sigma 0 to n on. Y^n/n! Which converges To e^Y in the case where Y=-1. Derived from Euler’s equation.

EricS
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