3

Find the following sum:

$$\sum_{k=1}^\infty \frac{1}{k}\frac{x^k}{k!}$$

where $x$ is a real number. This is a power series in $x$. In particular, I'm interested in the case $x>0$.

This is very similar to Calculate: $\sum_{k=1}^\infty \frac{1}{k^2}\frac{x^k}{k!}$, the only difference being the $k$ isn't squared in the denominator.

Disclaimer: This is not a homework exercise, I do not know if a closed form solution exists. If it doesn't, exist, then an approximation in terms of well-known functions (not the all-mighty general hypergeometric $_pF_q$, something simpler please) would be desired.

Community
  • 1
a06e
  • 6,665
  • WRA says a closed form exists: http://www.wolframalpha.com/input/?i=sum+x%5Ek%2F(k(k!))+from+k%3D1+to+infinity – Ethan Hunt Nov 09 '16 at 17:02
  • How do the answers to your other question not directly answer this one? You can use exactly the same methods. – Steven Stadnicki Nov 09 '16 at 17:44
  • 2
    It is clearly a hypergeometric function by the structure of its coefficients. And it is clearly an exponential integral, too. – Jack D'Aurizio Nov 09 '16 at 17:45
  • @StevenStadnicki As the answers here show, this sum turns out to be simpler than the one with the squared $k$. Over there the only choice seems to be an $_3F_3$. – a06e Nov 09 '16 at 18:06
  • @becko I'm not saying the answers are the same - I'm saying the techniques are exactly the same, and you should be able to use the answers to the other question to straightforwardly answer this one. I'm just a little confused as to what the purpose of this question was. – Steven Stadnicki Nov 09 '16 at 18:19

2 Answers2

4

Put $$f(x) = \sum_{k=1}^\infty \frac{1}{k} \frac{x^k}{k!}.$$ Then $$f'(x) = \sum_{k=1}^\infty \frac{x^{k-1}}{k!} = \frac{1}{x}\sum^\infty_{k=1} \frac{x^k}{k!} = \frac{e^x - 1}{x}.$$ Since $f(0) = 0$, we then get $$f(x) = \int^x_0 \frac{e^t - 1}{t} dt.$$ I'm fairly certain this is the best you'll get without resorting to special functions.

User8128
  • 15,485
  • 1
  • 18
  • 31
3

we have $$e^{x}-1=\sum _{k=1}^{\infty }{\frac {x^{k}}{k!}}$$ $$\frac{e^{x}-1}{x}=\sum _{k=1}^{\infty }{\frac {x^{k-1}}{k!}}$$

integrate both sides from $x=0$ to $x$ $$\int_{0}^{x}(\frac{e^{x}-1}{x})dx=\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}$$

so $$\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}=\int_{0}^{x}(\frac{e^{x}-1}{x})dx$$ $$\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}=Ei(x)-\log(x)-\gamma$$

E.H.E
  • 23,280