Question

The following series converge by the ratio test. Use summation by parts, $\sum _{k=1}^n{a}_k(b_{k+1}-b_k)=[a_{n+1}{b}_{n+1}-a_1{b}_1]-\sum _{k=1}^n{b}_{k+1}(a_{k+1}-a_k)$, to find the sum of the given series. $$\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}$$

Short answer

The sum of the series is $2$.

Step-by-step solution

Identify Terms

The series in question is $\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}$. Here, we can identify $a_n=n^2$ and $b_n=\frac{1}{2^n}$.

Perform Partial Sums

We want to find $\sum _{k=1}^n{a}_k(b_{k+1}-b_k)$. First, find $b_{k+1}-b_k=\frac{1}{2^{k+1}}-\frac{1}{2^k}=-\frac{1}{2^{k+1}}$. Thus, the expression becomes $\sum _{k=1}^n{k}^2(-\frac{1}{2^{k+1}})$.

Apply Summation by Parts Formula

According to the summation by parts formula, we have:$$\sum _{k=1}^n{a}_k(b_{k+1}-b_k)=[a_{n+1}{b}_{n+1}-a_1{b}_1]-\sum _{k=1}^n{b}_{k+1}(a_{k+1}-a_k)$$Substitute $a_k=k^2$ and $b_k=\frac{1}{2^k}$ into the formula.

Calculate Boundary Terms

Evaluate $a_{n+1}{b}_{n+1}=(n+1{)}^2\cdot \frac{1}{2^{n+1}}$ and $a_1{b}_1=1^2\cdot \frac{1}{2^1}=\frac{1}{2}$. This gives the boundary term as $\frac{(n+1{)}^2}{2^{n+1}}-\frac{1}{2}$.

Evaluate Remaining Sum

Determine $b_{k+1}=\frac{1}{2^{k+1}}$ and calculate $a_{k+1}-a_k=(k+1{)}^2-k^2=2k+1$. This simplifies the sum to $\sum _{k=1}^n\frac{2k+1}{2^{k+1}}$.

Conclude the Summation

Evaluate the full expression:$$\sum _{k=1}^n{a}_k(b_{k+1}-b_k)=[\frac{(n+1{)}^2}{2^{n+1}}-\frac{1}{2}]-\sum _{k=1}^n\frac{2k+1}{2^{k+1}}$$As $n\to \mathrm{\infty }$, the terms $\frac{(n+1{)}^2}{2^{n+1}}\to 0$. The simplified sum evaluates to an expression converging to a constant.

Key concepts

Ratio Test

The Ratio Test is a useful tool for determining the convergence or divergence of an infinite series. It works particularly well for series whose terms involve factorials or exponential functions. To apply the Ratio Test, you start by examining the terms of your series.
In our given series $\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}$, we need to determine if the following sequence:

• $a_n=\frac{n^2}{2^n}$
• $a_{n+1}=\frac{(n+1{)}^2}{2^{n+1}}$

is convergent by finding the limit $$L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|.$$Replacing into the formula, we have $$\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1{)}^2}{2^{n+1}}\times \frac{2^n}{n^2}=\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1{)}^2}{2n^2}=\frac{1}{2}.$$Since $\frac{1}{2}<1$, the Ratio Test tells us that the series is absolutely convergent. Thus, we have confirmed the series will converge.

Summation by Parts

Summation by parts is similar to integration by parts and provides a method to deal with sums that have two sequences multiplicatively connected. In our series problem, we use it to approach finding the exact sum of the series$$\sum _{n=1}^{\mathrm{\infty }}{a}_k(b_{k+1}-b_k)=[a_{n+1}{b}_{n+1}-a_1{b}_1]-\sum _{k=1}^n{b}_{k+1}(a_{k+1}-a_k).$$To apply this technique properly, first identify:

• $a_k=k^2$
• $b_k=\frac{1}{2^k}$

Then calculate:

• $b_{k+1}-b_k=\frac{1}{2^{k+1}}-\frac{1}{2^k}=-\frac{1}{2^{k+1}}$
• $a_{k+1}-a_k=(k+1{)}^2-k^2=2k+1$

By substituting these values, the formula helps simplify complex sums so that evaluating their convergence can be approached step-by-step. As shown in the solution, this method assists in rearranging and breaking down the series further for evaluation.

Infinite Series

An infinite series is essentially the sum of the terms of an infinite sequence. The notion of convergence is key here, as it determines whether adding up all these potentially endless terms comes to a finite figure. When we talk about convergence, we're interested if the sum approaches a particular value as you add more and more terms.In our problem, we are dealing with the infinite series $\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}$. Initially, it seemed challenging to calculate directly. However, by employing methods like the Ratio Test and Summation by Parts, we gain insight into its behavior and sum.The most important part is understanding that while each term becomes smaller as $n$ increases, the notion of reaching an exact sum hinges on these smaller terms diminishing rapidly enough for the whole series not to diverge to infinity. This involves methods to systematically verify and calculate sums, such as applying convergence tests and creative manipulation of terms, leading to evaluating whether an infinite series does converge to a particular constant.