$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ [duplicate]

Question

I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.

$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$

$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$

I was able to solve (1), as the integrand simplifies to $x^{e-2}$, however, I'm struggling with solving (2).

If we rewrite the roots as powers, we get:

$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$

combining the powers we get: $$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$

the exponent is the infinite sum

$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3} $$ we can split this into: $$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n} $$ The right sum is well known except here the sum begins at $n=1$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $x^3,x>0$ implying that (3) converges to 3 and the $\sum^{\infty}_{n=1}\frac{n}{2^n}$ converges to 2.

Which is part I'm struggling with. Any ideas?

$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$

Answer 1

Consider $$\sum_{n=1}^\infty n x^n=x\sum_{n=1}^\infty n x^{n-1}=x\left(\sum_{n=1}^\infty x^{n}\right)'$$

When finished, make $x=\frac 12$

Answer 2

Hint:

If $$f(x)=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $\forall$ $\vert x\vert \lt 1$. Then what is $$\frac{f'\left(\frac{1}{2}\right)}{2}$$

Answer 3

The binomial theoem gives$$(1-x)^{-2}=\sum_{n\ge0}(n+1)(-1)^nx^n,$$so the exponent is$$-1+(1-1/2)^{-2}=3.$$So the integral is $\int x^3dx=\frac14x^4+C$.