# Consider the following series \sum_{n=1}^{\infty}\frac{n}{2^{2n}} (a) Write the corresponding...

## Question:

Consider the following series {eq}\sum_{n=1}^{\infty}\frac{n}{2^{2n}} {/eq} (a) Write the corresponding improper integral for the series in the form {eq}\int_{1}^{\infty}f(x)dx {/eq} (b) What are the three properties of f(x) in order for the integral test to be applicable. (c) Evaluate the improper integral found in (a) Hint using integration by parts (d) Is the series convergent or divergent? Give reasons for your answer.

## Integral test:

The integral test is a test to determine if a series is convergent or divergent. We do this by integrating a function {eq}f {/eq} from the rule of the series and see if it converges with the limits. If the integral converges, so is the sum of the series.

## Answer and Explanation:

(a)

For the following series the improper integral is given by

$$\int _1^{\infty }\frac{x}{2^{2x}}dx \\$$

(b)

According to the integral test, {eq}f {/eq} should be continuous, positive, and decreasing function over the interval

(c)

Using integration by parts, let {eq}u = x{/eq} which means {eq}du = dx {/eq} and {eq}dv = \frac{1}{2^{2x}} dx {/eq} which means

$$v = \int \frac{1}{2^{2x}}dx \\$$

Let {eq}w = 2^{2x} {/eq} and {eq}dw = \ln \left(2\right)\cdot 2^{2x+1} dx {/eq}

\begin{align} v&= \int \frac{1}{2^{2x}}\times \frac{\ln \left(2\right)\cdot 2^{2x+1}}{\ln \left(2\right)\cdot 2^{2x+1}}dx \\ v &= \int \frac{1}{2^{2x}}\times \frac{1}{\ln \left(2\right)\cdot 2^{2x+1}}\times \ln \left(2\right)\cdot 2^{2x+1}dx \\ v &=\int \frac{1}{2^{2x}}\times \frac{1}{\ln \left(2\right)\cdot 2^{2x}\cdot 2}\times du \\ v &=\int \frac{1}{u}\times \frac{1}{\ln \left(2\right)\cdot u\cdot 2}\times du \\ v &=\frac{1}{ln\left(2\right)}\int \frac{1}{2u^2}du \\ v &=\frac{1}{2ln\left(2\right)}\int u^{-2}du \\ v &=\frac{1}{2ln\left(2\right)}\times -u^{-1} \\ v &=\frac{1}{2ln\left(2\right)}\times -\frac{1}{u} \\ v &=\frac{1}{2ln\left(2\right)}\times -\frac{1}{2^{2x}} \\ v &=\frac{1}{ln\left(2\right)}\times -\frac{1}{2^{2x+1}} \\ v &=-\frac{1}{2^{2x+1}ln\left(2\right)} \\ \end{align}

Going back to the integration by parts

\begin{align} \int _1^{\infty }\frac{x}{2^{2x}}dx &= uv - \int vdu \\ &=x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\int \left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)dx \\ &=x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)+\int \left(\frac{1}{2^{2x+1}ln\left(2\right)}\right)dx \\ \end{align}

Let {eq}w = 2^{2x} {/eq} which means {eq}dw = \ln \left(2\right)\cdot 2^{2x+1} dx {/eq}

\begin{align} &\int \frac{1}{2^{2x+1}ln\left(2\right)}\times \frac{2^{2x+1}ln\left(2\right)}{2^{2x+1}ln\left(2\right)}dx \\ &\int \frac{1}{2^{4x+2}ln^2\left(2\right)}\times 2^{2x+1}ln\left(2\right)dx \\ &\int \frac{1}{2^{4x}\times 2^2\times ln^2\left(2\right)}\times du \\ &\int \frac{1}{u^2\times 4\times ln^2\left(2\right)}\times du \\ &\frac{1}{4ln^2\left(2\right)}\int \frac{1}{u^2}du \\ &\frac{1}{4ln^2\left(2\right)}\times -\frac{1}{u} \\ &-\frac{1}{2^{2x}}\times \frac{1}{4ln^2\left(2\right)} \\ &-\frac{1}{2^{2x+2}ln^2\left(2\right)} \end{align}

Hence,

$$\int _1^{\infty }\frac{x}{2^{2x}}dx = x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\frac{1}{2^{2x+2}ln^2\left(2\right)} \\$$

Evaluating the limits of integration

\begin{align} &x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\frac{1}{2^{2x+2}ln^2\left(2\right)} \Biggr|_1^\infty \\ &\lim _{x\to \infty }\left[x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\frac{1}{2^{2x+2}ln^2\left(2\right)}\right]-\left[\left(1\right)\left(-\frac{1}{2^{2\left(1\right)+1}ln\left(2\right)}\right)-\frac{1}{2^{2\left(1\right)+2}ln^2\left(2\right)}\right] \\ &\lim _{x\to \infty }\left[x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\frac{1}{2^{2x+2}ln^2\left(2\right)}\right]-\left[-0.31042\right]=\left[0-0\right]-\left[-0.31042\right] \\ &\lim _{x\to \infty }\left[x\left(-\frac{1}{2^{2x+1}ln\left(2\right)}\right)-\frac{1}{2^{2x+2}ln^2\left(2\right)}\right]-\left[-0.31042\right]=0.31042 \\ \end{align}

$$\boxed{\int _1^{\infty }\frac{x}{2^{2x}}dx \approx 0.31042}$$

(d)

The series is convergent. Since the integral test is convergent so is the series.