Answer :
Let's analyze each series step-by-step to determine if they converge or diverge.
(a) [tex]\(\sum_{k=2}^{\infty} \frac{1}{k \ln(k) \ln(\ln(k))}\)[/tex]
To determine the convergence of this series, we can use the integral test. If the integral of the corresponding function converges, then the series converges, and if the integral diverges, the series diverges.
Consider the improper integral:
[tex]\[ \int_{2}^{\infty} \frac{1}{x \ln(x) \ln(\ln(x))} \, dx \][/tex]
Let's simplify this integral by making a substitution:
Let [tex]\( u = \ln(\ln(x)) \)[/tex], hence [tex]\( du = \frac{1}{\ln(x)} \cdot \frac{1}{x} dx \)[/tex].
Rewriting [tex]\( dx \)[/tex]:
[tex]\[ dx = x \ln(x) du \][/tex]
The integral becomes:
[tex]\[ \int_{u_2}^{\infty} \frac{1}{x \ln(x) u} x \ln(x) du \][/tex]
Simplifying:
[tex]\[ \int_{u_2}^{\infty} \frac{1}{u} du = \int_{\ln(\ln(2))}^{\infty} \frac{1}{u} du \][/tex]
This integral is a well-known divergent integral (the [tex]\( \ln(x) \)[/tex] integral diverges):
[tex]\[ \int_{\ln(\ln(2))}^{\infty} \frac{1}{u} du = \left. \ln|u| \right|_{\ln(\ln(2))}^{\infty} \][/tex]
As [tex]\( u \)[/tex] approaches infinity, [tex]\( \ln|u| \)[/tex] approaches infinity. So, the integral diverges, meaning the series:
[tex]\[ \sum_{k=2}^{\infty} \frac{1}{k \ln(k) \ln(\ln(k))} \][/tex]
diverges.
(b) [tex]\(\sum_{n=1}^{\infty} \frac{1}{n} \sin\left(\frac{n \pi}{2}\right)\)[/tex]
To determine the convergence of this series, let's analyze the behavior of [tex]\(\sin\left(\frac{n \pi}{2}\right)\)[/tex].
The function [tex]\(\sin\left(\frac{n \pi}{2}\right)\)[/tex] has a periodic pattern:
- When [tex]\( n = 1 \)[/tex], [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex], [tex]\( \sin(\pi) = 0 \)[/tex]
- When [tex]\( n = 3 \)[/tex], [tex]\( \sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- When [tex]\( n = 4 \)[/tex], [tex]\( \sin(2\pi) = 0 \)[/tex]
- When [tex]\( n = 5 \)[/tex], [tex]\( \sin\left(\frac{5\pi}{2}\right) = 1 \)[/tex]
- and so on...
Thus, the series can be written as:
[tex]\[ \frac{1}{1} \cdot 1 + \frac{1}{2} \cdot 0 + \frac{1}{3} \cdot (-1) + \frac{1}{4} \cdot 0 + \frac{1}{5} \cdot 1 + \cdots \][/tex]
So the series effectively reduces to:
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+1} - \sum_{n=0}^{\infty} \frac{1}{4n+3} \][/tex]
These are two Harmonic series with positive terms:
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+1} \][/tex]
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+3} \][/tex]
Both series individually diverge, leading to the conclusion that the original series diverges.
So, the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n} \sin\left(\frac{n \pi}{2}\right) \][/tex]
diverges.
(c) [tex]\(\sum_{i=0}^{\infty} \int_0^i e^{-x} dx\)[/tex]
First, compute the inner integral:
[tex]\[ \int_0^i e^{-x} \, dx \][/tex]
This integral evaluates to:
[tex]\[ \left[ -e^{-x} \right]_0^i = -e^{-i} + 1 = 1 - e^{-i} \][/tex]
Thus, the series transforms to:
[tex]\[ \sum_{i=0}^{\infty} \left(1 - e^{-i}\right) \][/tex]
We can split this sum into two separate sums:
[tex]\[ \sum_{i=0}^{\infty} 1 - \sum_{i=0}^{\infty} e^{-i} \][/tex]
The first sum is simply:
[tex]\[ \sum_{i=0}^{\infty} 1 \][/tex]
This series diverges because it is a sum of 1's. It's an infinite sum of a constant value:
[tex]\[ 1 + 1 + 1 + \cdots \][/tex]
The second series:
[tex]\[ \sum_{i=0}^{\infty} e^{-i} \][/tex]
is a geometric series with [tex]\( a = 1 \)[/tex] and [tex]\( r = e^{-1} \)[/tex], which converges to:
[tex]\[ \frac{1}{1 - e^{-1}} \][/tex]
However, since the diverging series [tex]\(\sum_{i=0}^{\infty} 1\)[/tex] is present, the entire series will diverge.
Thus, the series:
[tex]\[ \sum_{i=0}^{\infty} \int_0^i e^{-x} dx \][/tex]
diverges.
(a) [tex]\(\sum_{k=2}^{\infty} \frac{1}{k \ln(k) \ln(\ln(k))}\)[/tex]
To determine the convergence of this series, we can use the integral test. If the integral of the corresponding function converges, then the series converges, and if the integral diverges, the series diverges.
Consider the improper integral:
[tex]\[ \int_{2}^{\infty} \frac{1}{x \ln(x) \ln(\ln(x))} \, dx \][/tex]
Let's simplify this integral by making a substitution:
Let [tex]\( u = \ln(\ln(x)) \)[/tex], hence [tex]\( du = \frac{1}{\ln(x)} \cdot \frac{1}{x} dx \)[/tex].
Rewriting [tex]\( dx \)[/tex]:
[tex]\[ dx = x \ln(x) du \][/tex]
The integral becomes:
[tex]\[ \int_{u_2}^{\infty} \frac{1}{x \ln(x) u} x \ln(x) du \][/tex]
Simplifying:
[tex]\[ \int_{u_2}^{\infty} \frac{1}{u} du = \int_{\ln(\ln(2))}^{\infty} \frac{1}{u} du \][/tex]
This integral is a well-known divergent integral (the [tex]\( \ln(x) \)[/tex] integral diverges):
[tex]\[ \int_{\ln(\ln(2))}^{\infty} \frac{1}{u} du = \left. \ln|u| \right|_{\ln(\ln(2))}^{\infty} \][/tex]
As [tex]\( u \)[/tex] approaches infinity, [tex]\( \ln|u| \)[/tex] approaches infinity. So, the integral diverges, meaning the series:
[tex]\[ \sum_{k=2}^{\infty} \frac{1}{k \ln(k) \ln(\ln(k))} \][/tex]
diverges.
(b) [tex]\(\sum_{n=1}^{\infty} \frac{1}{n} \sin\left(\frac{n \pi}{2}\right)\)[/tex]
To determine the convergence of this series, let's analyze the behavior of [tex]\(\sin\left(\frac{n \pi}{2}\right)\)[/tex].
The function [tex]\(\sin\left(\frac{n \pi}{2}\right)\)[/tex] has a periodic pattern:
- When [tex]\( n = 1 \)[/tex], [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- When [tex]\( n = 2 \)[/tex], [tex]\( \sin(\pi) = 0 \)[/tex]
- When [tex]\( n = 3 \)[/tex], [tex]\( \sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- When [tex]\( n = 4 \)[/tex], [tex]\( \sin(2\pi) = 0 \)[/tex]
- When [tex]\( n = 5 \)[/tex], [tex]\( \sin\left(\frac{5\pi}{2}\right) = 1 \)[/tex]
- and so on...
Thus, the series can be written as:
[tex]\[ \frac{1}{1} \cdot 1 + \frac{1}{2} \cdot 0 + \frac{1}{3} \cdot (-1) + \frac{1}{4} \cdot 0 + \frac{1}{5} \cdot 1 + \cdots \][/tex]
So the series effectively reduces to:
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+1} - \sum_{n=0}^{\infty} \frac{1}{4n+3} \][/tex]
These are two Harmonic series with positive terms:
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+1} \][/tex]
[tex]\[ \sum_{n=0}^{\infty} \frac{1}{4n+3} \][/tex]
Both series individually diverge, leading to the conclusion that the original series diverges.
So, the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n} \sin\left(\frac{n \pi}{2}\right) \][/tex]
diverges.
(c) [tex]\(\sum_{i=0}^{\infty} \int_0^i e^{-x} dx\)[/tex]
First, compute the inner integral:
[tex]\[ \int_0^i e^{-x} \, dx \][/tex]
This integral evaluates to:
[tex]\[ \left[ -e^{-x} \right]_0^i = -e^{-i} + 1 = 1 - e^{-i} \][/tex]
Thus, the series transforms to:
[tex]\[ \sum_{i=0}^{\infty} \left(1 - e^{-i}\right) \][/tex]
We can split this sum into two separate sums:
[tex]\[ \sum_{i=0}^{\infty} 1 - \sum_{i=0}^{\infty} e^{-i} \][/tex]
The first sum is simply:
[tex]\[ \sum_{i=0}^{\infty} 1 \][/tex]
This series diverges because it is a sum of 1's. It's an infinite sum of a constant value:
[tex]\[ 1 + 1 + 1 + \cdots \][/tex]
The second series:
[tex]\[ \sum_{i=0}^{\infty} e^{-i} \][/tex]
is a geometric series with [tex]\( a = 1 \)[/tex] and [tex]\( r = e^{-1} \)[/tex], which converges to:
[tex]\[ \frac{1}{1 - e^{-1}} \][/tex]
However, since the diverging series [tex]\(\sum_{i=0}^{\infty} 1\)[/tex] is present, the entire series will diverge.
Thus, the series:
[tex]\[ \sum_{i=0}^{\infty} \int_0^i e^{-x} dx \][/tex]
diverges.
