Q. 86 - Calculus

Question

Prove Theorem 7.25. That is, show that the series $\sum _{k=1}^{\infty }{a}_k and \sum _{k=M}^{\infty }{a}_k$either both converge or both diverge. In addition, show that if $\sum _{k=M}^{\infty }{a}_k$converges to L, then $\sum _{k=1}^{\infty }{a}_k$converges to $a_1+a_2+a_3+....+a_{M-1}+L.$

Step-by-Step Solution

Answer

It is shown that the series $\sum _{k=1}^{\infty }{a}_k and \sum _{k=M}^{\infty }{a}_k$ either both converges or both diverges.

Also, it is shown if $\sum _{k=M}^{\infty }{a}_k$ converges to L, then $\sum _{k=1}^{\infty }{a}_k$ converges to  $a_1+a_2+a_3+....+a_{M-1}+L.$

Step 1. Given Information.

We are given two series: $\sum _{k=1}^{\infty }{a}_k$and $\sum _{k=M}^{\infty }{a}_k$.

We need to show that either both the series converges or both diverge.

And also if $\sum _{k=M}^{\infty }{a}_k$ converges to $L$, then $\sum _{k=1}^{\infty }{a}_k$ converges to $a_1+a_2+a_3+....+a_{M-1}+L.$

Step 2. Proof of the first part

Let us assume that the series $\sum _{k=1}^{\infty }{a}_k$ is convergent. Then we have to show that the series $\sum _{k=M}^{\infty }{a}_k$ is also convergent.

The series $\sum _{k=1}^{\infty }{a}_k$ is convergent and converges to $A$. Therefore, the sequence $\left\{A_n\right\}$ of the partial sum of the series $\sum _{k=1}^{\infty }{a}_k$is convergent and converges to $A$.

Let the sequence $\left\{B_n\right\}$ be the sequence of the partial sum of the series $\sum _{k=M}^{\infty }{a}_k$.

Therefore, by the definition of convergence of a sequence, for given $\epsilon >0$, there exist a positive integer $N$ such that $\left|A_n-A\right|<\epsilon$ for $k\ge N$......(1)

For $k\ge M$, the terms of the sequences $\left\{A_n\right\}$ and $\left\{B_n\right\}$ are the same.

Choose $P=max\left\{N,M\right\}$.

Therefore, $\left|B_n-A\right|<\epsilon$ for $k\ge M$ ......(2)

Therefore, for given $\epsilon >0$ there exist a positive integer P such that $\left|B_n-A\right|<\epsilon$ for $k\ge M$.

Hence, the sequence $\left\{B_n\right\}$ of partial sum of the series $\sum _{k=M}^{\infty }{a}_k$ is convergent and converges to $A$.

Thus, the series $\sum _{k=M}^{\infty }{a}_k$ is convergent.

Step 3. Proof for the second part

It is given that the series $\sum _{k=M}^{\infty }{a}_k$ is convergent and converges to $L$.

We need to show that the series $\sum _{k=1}^{\infty }{a}_k$ converges to $a_1+a_2+a_3+...+a_{M-1}+L$.

The series $\sum _{k=1}^{\infty }{a}_k$ can be written as

$$\begin{gathered} \sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{M-1}{a}_k+\sum _{k=M}^{\infty }{a}_k \\ \sum _{k=1}^{\infty }{a}_k=a_1+a_2+a_3+...+a_{M-1}+\sum _{k=M}^{\infty }{a}_k \\ \sum _{k=1}^{\infty }{a}_k=a_1+a_2+a_3+...+a_{M-1}+L \end{gathered}$$

So the series converges to $a_1+a_2+a_3+...+a_{M-1}+L$.