0

Please help me to solve this problem. I need to find the sum of an infinite series: $$ S = 1 + 1 + \frac34 + \frac12 + \frac5{16} + \cdots $$

I tried to imagine this series as a derivative of a certain function $$ f(x) = x + 2x^{1/2} + 3x^{1/4} + 4x^{1/8} + \cdots + nx^{1/2^{n-1}}. $$

But I did not understand how to find the value of the derivative at the point $x = 1$.

I would be very grateful if someone would help me. Thank you in advance.

Sammy Black
  • 28,409
Starnex
  • 21
  • 2
    Derivates of fractional exponents won't work. Try to write out the series formally, first. – Thomas Andrews May 13 '24 at 18:05
  • 3
    Instead of these fractional exponents, in order to get these powers of $2$, imagine that your power series has $1/2$ substituted into it. Then, the $a_n x^n$ term will become $a_n / 2^n$, which might make the series easier to identify. – Theo Bendit May 13 '24 at 18:09
  • 2
    Hi! Wouldn't it be clearer if you wrote this series as $1 + \frac22 + \frac34 + \frac48 + \frac{5}{16}+\cdots$? I appreciate that you've reduced the fractions, but in this case that tends to obscure the pattern without really getting closer to the solution. That also makes it easier to check whether you've accurately done what @ThomasAndrews suggests. – Brian Tung May 13 '24 at 18:37
  • 1
    Also by reducing your fractions, you just make it harder for people to figure out what the series is. You are asking people to volunteer their time to help you, don't waste their time initially. – Thomas Andrews May 13 '24 at 18:56
  • @ThomasAndrews: Indeed, I really meant "clearer to readers", but it's certainly true that it should make it clearer to OP as well. ETA: I don't think they meant to waste readers' time, I think they thought it would be helpful. But it doesn't really help. – Brian Tung May 13 '24 at 19:03
  • 1
    Oh, I don't think they meant to waste anybody's time, either, but that doesn't mean they didn't. @BrianTung – Thomas Andrews May 13 '24 at 19:14

1 Answers1

1

Notice that your series is equal to

$$\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3} +\dots=\sum^{\infty}_{n=0}\frac{n}{2^{n-1}}=\sum^{\infty}_{n=0}2^{1-n}n=2 \sum^{\infty}_{n=0}2^{-n}n$$

Let's check if this series converges

Take the limit as n approaches $\infty$ of $\left| \frac{\frac{n+1}{2^{(n+1)-1}}}{\frac{n}{2^{n-1}}}\right|$

$$\lim _{n \to \infty} \left| \frac{\frac{n+1}{2^{(n+1)-1}}}{\frac{n}{2^{n-1}}}\right|=\frac{1}{2}$$

Since the limit is less than 1 , the series converges absolutely .

Consider the sum $$\sum ^{\infty}_{n=0} Z_0^n=\frac{1}{1-Z_0}$$

Differentiate both sides with respect to $Z_0$

$$\sum ^{\infty}_{n=0}\frac{nZ^n_0}{Z_0}=\frac{1}{(Z_0-1)^2}$$

Multiply by $Z_0$

$$\sum ^{\infty}_{n=0}nZ^n_0=\frac{Z_0}{(Z_0-1)^2} $$

Setting $Z_0=\frac{1}{2}$ , we get that

$$\sum ^{\infty}_{n=0}2^{-n}n=2$$

therefore ,

$$2\left(\sum ^{\infty}_{n=0}2^{-n}n\right )=2(2)$$

Hence ,

$$\sum ^{\infty}_{n=0}2^{-n}n=4$$

Antony Theo.
  • 4,716