Logarithm of an infinite series I am curious about simplifying the following expression:
$$
\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v}, v>0, x>0 
$$
Is there any rule to simplify a summation inside the log? 
Also, is there any way to derive the following expression:
$$
\frac{d}{dx}\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v} 
$$
or
$$
\frac{d}{dv}\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v} 
$$
Thanks!
 A: The exponential function $e^y$ is defined as 
$$ e^y:=\sum_{k = 0}^{\infty} {y^k \over k!}.$$ 
So your sum is simply (by setting $y=x^{\frac{1}{v}}$)
$$e^{x^{\frac{1}{v}}}.$$
Summary


*

*If you mean to calculate the logarithm of the sum raised to the power of $v$, log(sum^v):


$$\begin{align}
\log \left(\left(\sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)^v\right)&=v\log \left(\sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)\\
&=v\log\left(e^{x^{\frac{1}{v}}}\right)\\
&=vx^{\frac{1}{v}}\log e \\
&=vx^{\frac{1}{v}}.\end{align}$$


*

*If you mean to calculate the logarithm of the sum all raised to the power of $v$, (log(sum))^v:


$$\begin{align}
\left(\log \sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)^v&=\left(\log e^{x^{\frac{1}{v}}}\right)^v\\
&=\left(x^{\frac{1}{v}}\right)^v\\
&=x^{\frac{v}{v}}=x.\end{align}$$
A: The exponential function is defined by
$$e^z = \sum\limits_{n=0}^\infty \frac{z^n}{n!}.$$
Hence we have that
$$\log\left(\sum\limits_{n=0}^\infty \frac{x^{n/v}}{n!}\right)^v = \log\left(\sum\limits_{n=0}^\infty \frac{(x^{1/v})^n}{n!}\right)^v = \log\left(e^{x^{1/v}}\right)^v = (x^{1/v})^v = x. $$
Now that we've simplified the original expression, it should be easy to compute the derivatives.
A: Let's start over. As the other answers already worked it out, the expression inside the logarithm, or rather inside the big parentheses, simplifies because it's an instance of the series for the exponential function:
$$\sum_{n=0}^{\infty}\frac{x^{\frac{n}{v}}}{n!}=\sum_{n=0}^{\infty}\frac{\left(x^{1/v}\right)^n}{n!}=e^{x^{1/v}}.$$
But then remember what $a^{b^c}$ means: $a^{b^c}=a^{(b^c)}$, not $(a^b)^c$ (which would be equal to $a^{bc}$). So we get
$$\log\left(\sum_{n=0}^{\infty}\frac{x^{\frac{n}{v}}}{n!}\right)^v=\log\left(e^{x^{1/v}}\right)^v=v\log\left(e^{x^{1/v}}\right)=vx^{1/v}.$$
From that, you can take the derivative either with respect to $x$ or with respect to $y$. Let us know if you'd like to see that worked out.
A: We know that $$ e^{x} =\sum_{k =0}^{\infty} \frac {x^k}{k!}$$ In our problem we can write our expression as $$\log(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}}{n!})^v =\log(\sum_{n=0}^{\infty} \frac {(x^{\frac {1}{v}})^n}{n!})^v$$ Thus we have, $$\log(e^{x^{\frac {1}{v}}})^v = (x^{\frac {1}{v}})^v=x$$ Hope it helps. 
