Harmonic and Something Else

Relevant wiki: Maclaurin Series

Considering the Maclaurin series of

$\begin{aligned} \ln(1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \cdots & \small \color{#3D99F6} \text{for }-1 < x \le 1 \\ \ln(1-x) & = - \sum_{n=1}^\infty \frac {x^n}n \\ \implies \sum_{n=1}^\infty \frac {x^n}n & = - \ln (1-x) = \ln \left(\frac 1{1-x}\right) & \small \color{#3D99F6} \text{Putting }x=\frac 12 \\ \sum_{n=1}^\infty \frac 1{n2^n} & = \ln \left(\frac 1{1-\frac 12}\right) = \boxed{\ln 2} \end{aligned}$

First, compare the series with the infinite geometric series,

$\displaystyle \dfrac1{2} + \dfrac1{4} + \dfrac1{8} + \dfrac1{16} + \cdots = 1$ $\displaystyle \dfrac1{1*2} + \dfrac1{2*4} + \dfrac1{3*8} + \dfrac1{4*16} + \cdots$

Clearly, every term in the bottom series is equal to or less than its term in the top series. Therefore, the sum we are looking for is less than 1.

Then, note that $\dfrac1{e}$ is less than $\dfrac1{2}$ since $2$ < $e$ . Since the first term in the series is $\dfrac1{2}$ and all the other terms are positive, the series converges to something greater than $\dfrac1{e}$ . The only option left among the possible answer choices is $\ln(2)$ so this must be the answer.

Let $|x| < 1$ and $y = \sum_{n = 1}^{\infty} \dfrac{x^n}{n}$

$\implies \dfrac{dy}{dx} = \sum_{n = 1}^{\infty} x^{n - 1} = \dfrac{1}{1 - x} \implies y = \int_{0}^{x} \dfrac{1}{1 - x} dx = \ln|\dfrac{1}{1 - x}|$ .

Using $x = \dfrac{1}{2} \implies$ $\sum_{n=1}^{\infty}\dfrac1{n2^n} = \boxed{\ln(2)}$ .

Share my solution:

Let $\displaystyle f(x)=\frac x2+\frac{x^2}{2\cdot2^2}+\frac{x^3}{3\cdot2^3}+\dots+\frac{x^k}{k\cdot2^k}+\dots$ , so question is asking $f(1)$ .

Doing derivative, get

$\begin{aligned}f'(x)&=\frac12+\frac x{2^2}+\frac{x^2}{2^3}+\frac{x^3}{2^4}+\dots\\ &=\frac{1/2}{1-x/2}\\ &=\frac1{2-x}\end{aligned}$

Note that the function only converge when $-2<x<2$ which didn't contradict our assumption.

So, take anti-derivative get

$\begin{aligned}f(x)&=\int f'(x)dx\\ &=\int \frac1{2-x}dx\\ &=-\ln{|2-x|}+C\end{aligned}$

Look back to the expression of $f(x)$ , we know that $f(0)=0$ , leads to $C=\ln2$ , so answer would be:

$f(1)=-\ln{|2-1|}+\ln2=\boxed{\ln2}$

Many great solutions were given, and this is a way to check your answer in Python:

sum([1/(n*2**n) for n in range(1, 100)])

we'll write $f(x)=\frac{x}{1}+\frac{x^2}{2}+\dots=\sum_{n=1}^{\infty}\frac{x^n}{n}$

then, we will differentiate to get a geometric series $f'(x)=1+x+x^2+\dots=\sum_{n=1}^{\infty}x^n=\frac{1}{1-x}$

now we can integrate to get our equation back $f'(x)=1+x+x^2+\dots=\sum_{n=1}^{\infty}x^n=\frac{1}{1-x}\\u=1-x\\du=-dx\\f(x)=-\int\frac{1}{u}du=-\ln{u}=-\ln{(1-x)}=\ln{\frac{1}{1-x}}$

we need to calculate $f(\frac{1}{2})$ , which is $\ln{\frac{1}{1-\frac{1}{2}}}=\ln{\frac{2}{2-1}}=\ln{2}$

As we have to add (1÷2) in the converging infinite series the answer is between 1 and 0.5 so the answer is ln2

This can be done through process of elimination.

$\displaystyle \frac{1}{\left(\sqrt{2}\right)^3 - 1} = \sum_{n=1}^\infty \frac{1}{\left(\sqrt{2}\right)^n 2^n} < \sum_{n=1}^\infty \frac{1}{n2^n}< \sum_{n=1}^\infty \frac{1}{2^n} = 1$

Since $\frac{1}{\left(\sqrt{2}\right)^3 - 1} \approx 0.547$ and $\frac{1}{e}\approx 0.368$

the answer must be $\ln 2$