Question

Find the sum of the series: $\frac{1}{2}+\frac{1}{8}+\frac{1}{24}+\frac{1}{64}+......+\frac{1}{k.2^{k}}+......$

Answer 1

ln 2

Answer 2

The given series is $S = \frac{1}{2}+\frac{1}{8}+\frac{1}{24}+\frac{1}{64}+......$.
The general term of the series can be identified as $T_k = \frac{1}{k \cdot 2^k}$.
So, the series can be written in summation form as:
$S = \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^k}$

We can rewrite this as:
$S = \sum_{k=1}^{\infty} \frac{(1/2)^k}{k}$

Recall the Maclaurin series expansion for $\ln(1-x)$:
The Maclaurin series for $\ln(1-x)$ is given by:
$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$
This can be expressed in summation notation as:
$\ln(1-x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}$
This expansion is valid for $|x| < 1$.

Comparing our series $S = \sum_{k=1}^{\infty} \frac{(1/2)^k}{k}$ with the general form $-\ln(1-x) = \sum_{k=1}^{\infty} \frac{x^k}{k}$, we can see that $x = \frac{1}{2}$.

Since $|x| = |1/2| = 1/2$, which is less than 1, the expansion is valid.
Substitute $x = 1/2$ into the formula for $\ln(1-x)$:
$\sum_{k=1}^{\infty} \frac{(1/2)^k}{k} = -\ln\left(1 - \frac{1}{2}\right)$
$S = -\ln\left(\frac{1}{2}\right)$
Using the logarithm property $\ln(a/b) = \ln a - \ln b$ and $\ln(1) = 0$:
$S = -(\ln 1 - \ln 2)$
$S = -(0 - \ln 2)$
$S = \ln 2$

The sum of the series is $\ln 2$.