In Class Practice

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n$

$S_k=\sum _{n=1}^k(\frac{1}{2}{)}^k$

First prove that $S_n=1-\frac{1}{2^n}$

Proof. Let $n\in \mathbb{N}$ and $P(n):\sum _{k=1}^n(\frac{1}{2}{)}^k=1-\frac{1}{2^n}$. We work by induction to show that $P(n)$ is true for all $n\in \mathbb{N}$.

Base case. Note that $\sum _{k^1{\left(\frac{1}{2}\right)}^k}=\frac{1}{2}=1-{\frac{1}{2}}^1$. Thus $P(1)$ is true.
Inductive step. Let $n\in \mathbb{N}$ and assume that $P(n)$ is true. Then $\sum _{k=1}^n{\left(\frac{1}{2}\right)}^k=1-{\frac{1}{2}}^n$. Then $\sum _{k=1}^{n+1}(\frac{1}{2}{)}^k=\sum _{k=1}^n(\frac{1}{2}{)}^k+(\frac{1}{2}{)}^{n+1}$. Then by the inductive hypothesis $=1-{\frac{1}{2}}^n+{\frac{1}{2}}^{n+1}=1-{\left(\frac{1}{2}\right)}^{n+1}$. Thus by principle of mathematical induction, $P(n)$ holds true for all $n\in \mathbb{N}$.

Now prove that $li{m}_{n\to \mathrm{\infty }}{S}_n=1$.

Scratch work.
$\epsilon >0$

Proof.
For $n\in \mathbb{N}$, define $S_n=1-{\frac{1}{2}}^n$. We show that $\underset{n\to \mathrm{\infty }}{lim}{S}_n=1$. Let $\epsilon >0$. Fix $N={\log }_{\frac{1}{2}}(\epsilon )$. Let $n\in \mathbb{N},n>N$. Then

Thus $\underset{n\to \mathrm{\infty }}{lim}{S}_n=1$. $◻$

Theorem: Let $(a_n)$ be a sequence of real numbers.

HW 34

Ben Finch

E 34.1

Proof.
We will prove that $s_n=\frac{n(n+1)}{2}$. Let $n\in \mathbb{N}$ and $P(n):\frac{n(n+1)}{2}$. We work by induction to show $P(n)$ is true for all $n\in \mathbb{N}$.

Base Case:
Note that $P(1)=\frac{1(1+1)}{2}=1$. So $P(1)$ is true.

Inductive step:
Let $k\in \mathbb{N}$ and assume $P(k)$ is true. Then $P(k)=\frac{k(k+1)}{2}$. So

Thus by principle of mathematical induction, $P(n)$ holds true for all $n\in \mathbb{N}$.

E 34.2
Proof.

Denote the $n_{th}$ partial sum as $s_n$ and establish $P(n):s_n=\frac{n}{2}[2c+(n-1)d]$. We need to show that $P(n)$ is true for all $n\in \mathbb{N}$.

Base Case:
Consider $n=1$. Then $s_1=a_1=c$, which matches the formula:
$P(1)=\frac{1}{2}[2c+(1-1)d]=c.$
So $P(1)$ is true.

Inductive Step:
Assume $P(k)$ is true for some $k\in \mathbb{N}$, i.e., $s_k=\frac{k}{2}[2c+(k-1)d]$. We need to show that $P(k+1)$ is also true.

Now, $s_{k+1}=s_k+a_{k+1}$. By the inductive hypothesis, $s_k=\frac{k}{2}[2c+(k-1)d]$, and $a_{k+1}=c+kd$.

Hence,

Thus, $P(k+1)$ holds.

By the principle of mathematical induction, $P(n)$ holds true for all $n\in \mathbb{N}$. $◻$

E 34.3

a)
We first prove that $S_n=1-\frac{1}{2^n}$

Proof. Let $n\in \mathbb{N}$ and $P(n):\sum _{k=1}^n(\frac{1}{2}{)}^k=1-\frac{1}{2^n}$. We work by induction to show that $P(n)$ is true for all $n\in \mathbb{N}$.

Base case. Note that $\sum _{k^1{\left(\frac{1}{2}\right)}^k}=\frac{1}{2}=1-{\frac{1}{2}}^1$. Thus $P(1)$ is true.
Inductive step. Let $n\in \mathbb{N}$ and assume that $P(n)$ is true. Then $\sum _{k=1}^n{\left(\frac{1}{2}\right)}^k=1-{\frac{1}{2}}^n$. Then $\sum _{k=1}^{n+1}(\frac{1}{2}{)}^k=\sum _{k=1}^n(\frac{1}{2}{)}^k+(\frac{1}{2}{)}^{n+1}$. Then by the inductive hypothesis $=1-{\frac{1}{2}}^n+{\frac{1}{2}}^{n+1}=1-{\left(\frac{1}{2}\right)}^{n+1}$. Thus by principle of mathematical induction, $P(n)$ holds true for all $n\in \mathbb{N}$. $◻$

b)
We now prove that $li{m}_{n\to \mathrm{\infty }}{S}_n=1$.

Scratch work.
$\epsilon >0$

Proof.
For $n\in \mathbb{N}$, define $S_n=1-{\frac{1}{2}}^n$. We show that $\underset{n\to \mathrm{\infty }}{lim}{S}_n=1$. Let $\epsilon >0$. Fix $N={\log }_{\frac{1}{2}}(\epsilon )$. Let $n\in \mathbb{N},n>N$. Then

Thus $\underset{n\to \mathrm{\infty }}{lim}{S}_n=1$. $◻$

E 34.4
We aim to prove that $\underset{n}{lim}\to \mathrm{\infty }\frac{1}{3^n}=0$.

Scratch work.

Proof. Let $\epsilon >0$. Fix $N=\frac{\ln \left(\frac{1}{\epsilon }\right)}{\ln (3)}$. Let $n>N,n\in \mathbb{N}$. Then

Thus, $\epsilon >|\frac{1}{3^n}-0|$ and $\underset{n}{lim}\to \mathrm{\infty }\frac{1}{3^n}=0$. Therefore, by Theorem 34.6, the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3^n}$converges. $◻$

E 34.5
a)
Consider the definition of $t_n$:

This sum includes $2^n-(2^{n-1}+1)+1=2^n-2^{n-1}$ terms, which is $2^{n-1}$ terms. Now, the largest term in this sum is $\frac{1}{2^{n-1}+1}$, and the smallest term is $\frac{1}{2^n}$.

Since each term $\frac{1}{k}$ in the sum is at least as large as $\frac{1}{2^n}$, we can estimate the sum by replacing each term with $\frac{1}{2^n}$:

So, each partial sum $t_n$ is at least $\frac{1}{2}$, and this lower bound doesn't depend on $n$; it is the same for all $n\ge 1$.

Therefore, we have proved that $t_n\ge \frac{1}{2}$ for all $n\ge 1$. $◻$

b)
Proof.

Base Case:

For $n=1$, $s_{2^1}$ is the sum of the first two terms of the harmonic series:

$s_2=1+\frac{1}{2}\ge 1+\frac{1}{2}$

This is true, so the base case is verified for $n=1$.

Inductive step:

Assume that the statement is true for some $k\ge 1$:

$s_{2k}\ge 1+\frac{k}{2}$

We want to show that the statement holds for $k+1$:

$s_{2(k+1)}=s_{2k+2}\ge 1+\frac{k+1}{2}$

Using the hint, we know that:

$s_{2n+1}=s_{2n}+t_{n+1}$

and from part (a), we have $t_n\ge \frac{1}{2}$ for each $n\ge 1$.

Therefore, we can write:

$s_{2k+2}=s_{2k+1}+\frac{1}{2k+2}$

$=(s_{2k}+t_{k+1})+\frac{1}{2k+2}$

By the induction hypothesis, $s_{2k}\ge 1+\frac{k}{2}$, and we've shown that $t_{k+1}\ge \frac{1}{2}$, so:

$s_{2k+2}\ge (1+\frac{k}{2})+\frac{1}{2}+\frac{1}{2k+2}$

Thus, by induction, the statement $s_{2n}\ge 1+\frac{n}{2}$ is true for all $n\ge 1$. $◻$

c)
Proof. We aim to prove that the harmonic series does not converge.

We need to show that for any $L\in \mathbb{R}$, there exists an $\epsilon >0$ such that for any $N\in \mathbb{N}$, we can find an $n\in \mathbb{N}$, $n>N$, for which $|s_n-L|\ge \epsilon$.

Choose $\epsilon =\frac{1}{2}$. Now, we want to show that for any $N\in \mathbb{N}$, we can find an $n\in \mathbb{N}$, $n>N$, such that $|s_n-L|\ge \frac{1}{2}$.

From part (a), we know that each $t_n$ in the harmonic series, defined as $t_n=\sum _{k=2^{n-1}+1}^{2^n}\frac{1}{k}$, is at least $\frac{1}{2}$.

Given any $N\in \mathbb{N}$, let $n$ be such that $2^n>N$. This means $n>{\log }_2(N)$. The partial sum $s_{2^n}$ includes the term $t_n$, which is at least $\frac{1}{2}$. Therefore, for any $L\in \mathbb{R}$, we have:

$|s_{2^n}-L|\ge |t_n|\ge \frac{1}{2}$

Thus, we have found an $n$ for which the positive difference between the $n$-th partial sum of the harmonic series and any real number $L$ is at least $\frac{1}{2}$, which means that the sequence of partial sums does not converge to $L$. Thus the harmonic sequence does not converge. $◻$