Question(10) The sequences $a_{n}=\frac{e^{n}+e^{-n}}{e^{2 n}+1}$ (a) converges to 0 (b) converges to 1 (c) converges to $\frac{1}{2}$ (d) diverges (11) $\int_{2}^{\infty} \frac{d x}{\sqrt{x^{3}-1}}$ (a) converges by direct comparison with $\int_{2}^{\infty} \frac{d x}{x^{3 / 2}}$ (b) diverges by direct comparison with $\int_{2}^{\infty} \frac{d x}{x^{3}}$ (c) converges by limit comparison with $\int_{2}^{\infty} \frac{d x}{x^{3 / 2}}$ (d) diverges by limit comparison with $\int_{2}^{\infty} \frac{d x}{x^{3}}$ (12) The series $\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right)$ (a) converges to $\ln 2$ (b) diverges by L.C.T. with $\sum_{n=1}^{\infty} \frac{1}{n}$ (c) converges L.C.T. with $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ (d) divegres by D.C.T with $\sum_{n=1}^{\infty} \frac{1}{n}$ (13) $\int_{3}^{\infty} \frac{\ln ^{2} x}{x^{3}} d x$ (a) converges by direct comparison with $\int_{3}^{\infty} \frac{d x}{x^{3}}$ (b) diverges by direct comparison with $\int_{3}^{\infty} \frac{d x}{x}$ C. converges by direct comparison with $\int_{3}^{\infty} \frac{d x}{x^{2}}$ (d) diverges by direct comparison with $\int_{3}^{\infty} \frac{d x}{\sqrt{x}}$

Studdy Solution

STEP 1

What is this asking? We've got a fun mix here!
We need to figure out if a sequence converges, decide whether two integrals converge or diverge, and determine the convergence or divergence of a series. Watch out! Don't get tricked by the looks of these problems!
Make sure to use the right convergence tests and be careful with your comparisons.

STEP 2

1. Analyze the sequence
2. Tackle the first integral
3. Investigate the series
4. Conquer the second integral

STEP 3

Alright, let's rewrite our sequence $a_n = \frac{e^n + e^{-n}}{e^{2n} + 1}$ to make it easier to work with.
We can divide both the numerator and the denominator by $e^{2n}$ : $\frac{\frac{e^n}{e^{2n}} + \frac{e^{-n}}{e^{2n}}}{\frac{e^{2n}}{e^{2n}} + \frac{1}{e^{2n}}} = \frac{e^{-n} + e^{-3n}}{1 + e^{-2n}}.$ Why did we do this?
It helps us see what happens when $n$ gets really large!

STEP 4

As $n$ approaches infinity, $e^{-n}$ , $e^{-2n}$ , and $e^{-3n}$ all approach zero.
So, our sequence becomes: $\lim_{n \to \infty} \frac{e^{-n} + e^{-3n}}{1 + e^{-2n}} = \frac{0 + 0}{1 + 0} = \frac{0}{1} = 0.$ So, the sequence converges to 0!

STEP 5

We're looking at $\int_{2}^{\infty} \frac{dx}{\sqrt{x^3 - 1}}$ .
For large $x$ , $\sqrt{x^3 - 1}$ behaves like $\sqrt{x^3} = x^{3/2}$ .
Let's use the Limit Comparison Test with $\int_{2}^{\infty} \frac{dx}{x^{3/2}}$ .

STEP 6

We need to evaluate the limit: $\lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^3 - 1}}}{\frac{1}{x^{3/2}}} = \lim_{x \to \infty} \frac{x^{3/2}}{\sqrt{x^3 - 1}} = \lim_{x \to \infty} \sqrt{\frac{x^3}{x^3 - 1}} = \lim_{x \to \infty} \sqrt{\frac{1}{1 - \frac{1}{x^3}}} = \sqrt{\frac{1}{1 - 0}} = 1.$ Since the limit is a finite positive number (1), both integrals behave the same way.

STEP 7

We know that $\int_{2}^{\infty} \frac{dx}{x^{3/2}}$ converges because it's a p-integral with $p = \frac{3}{2} > 1$ .
Therefore, our original integral also converges!

STEP 8

We have the series $\sum_{n=1}^{\infty} \ln\left(1 + \frac{1}{n^2}\right)$ .
Remember that for small $x$ , $\ln(1 + x) \approx x$ .
So, for large $n$ , $\ln\left(1 + \frac{1}{n^2}\right) \approx \frac{1}{n^2}$ .
Let's use the Limit Comparison Test with $\sum_{n=1}^{\infty} \frac{1}{n^2}$ .

STEP 9

We evaluate the limit: $\lim_{n \to \infty} \frac{\ln\left(1 + \frac{1}{n^2}\right)}{\frac{1}{n^2}}$ Let $u = \frac{1}{n^2}$ .
As $n \to \infty$ , $u \to 0$ .
So, the limit becomes: $\lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1.$ Since the limit is a finite positive number (1), both series behave the same way.

STEP 10

Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ ), our original series also converges!

STEP 11

We're dealing with $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ .
We know that for $x > 1$ , $0 < \ln x < x$ .
Let's square this inequality: $0 < \ln^2 x < x^2$ .

STEP 12

Now, divide by $x^3$ : $0 < \frac{\ln^2 x}{x^3} < \frac{x^2}{x^3} = \frac{1}{x}$ .
This doesn't help us much, since $\int_{3}^{\infty} \frac{dx}{x}$ diverges.

STEP 13

Instead, we can use that for large $x$ , $\ln x$ grows much slower than any positive power of $x$ .
So, for sufficiently large $x$ , $\ln^2 x < x$ .
Therefore, $\frac{\ln^2 x}{x^3} < \frac{x}{x^3} = \frac{1}{x^2}$ .

STEP 14

Since $\int_{3}^{\infty} \frac{dx}{x^2}$ is a convergent p-integral ( $p = 2 > 1$ ), by the Direct Comparison Test, our original integral $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ also converges!

STEP 15

(10) (a) (11) (c) (12) (c) (13) (c)

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Studdy Solution

STEP 1

STEP 2

1. Analyze the sequence
2. Tackle the first integral
3. Investigate the series
4. Conquer the second integral

STEP 3

STEP 4

As $n$ approaches infinity, $e^{-n}$ , $e^{-2n}$ , and $e^{-3n}$ all approach zero.
So, our sequence becomes: $\lim_{n \to \infty} \frac{e^{-n} + e^{-3n}}{1 + e^{-2n}} = \frac{0 + 0}{1 + 0} = \frac{0}{1} = 0.$ So, the sequence converges to 0!

STEP 5

We're looking at $\int_{2}^{\infty} \frac{dx}{\sqrt{x^3 - 1}}$ .
For large $x$ , $\sqrt{x^3 - 1}$ behaves like $\sqrt{x^3} = x^{3/2}$ .
Let's use the Limit Comparison Test with $\int_{2}^{\infty} \frac{dx}{x^{3/2}}$ .

STEP 6

STEP 7

We know that $\int_{2}^{\infty} \frac{dx}{x^{3/2}}$ converges because it's a p-integral with $p = \frac{3}{2} > 1$ .
Therefore, our original integral also converges!

STEP 8

We have the series $\sum_{n=1}^{\infty} \ln\left(1 + \frac{1}{n^2}\right)$ .
Remember that for small $x$ , $\ln(1 + x) \approx x$ .
So, for large $n$ , $\ln\left(1 + \frac{1}{n^2}\right) \approx \frac{1}{n^2}$ .
Let's use the Limit Comparison Test with $\sum_{n=1}^{\infty} \frac{1}{n^2}$ .

STEP 9

STEP 10

Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ ), our original series also converges!

STEP 11

We're dealing with $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ .
We know that for $x > 1$ , $0 < \ln x < x$ .
Let's square this inequality: $0 < \ln^2 x < x^2$ .

STEP 12

Now, divide by $x^3$ : $0 < \frac{\ln^2 x}{x^3} < \frac{x^2}{x^3} = \frac{1}{x}$ .
This doesn't help us much, since $\int_{3}^{\infty} \frac{dx}{x}$ diverges.

STEP 13

Instead, we can use that for large $x$ , $\ln x$ grows much slower than any positive power of $x$ .
So, for sufficiently large $x$ , $\ln^2 x < x$ .
Therefore, $\frac{\ln^2 x}{x^3} < \frac{x}{x^3} = \frac{1}{x^2}$ .

STEP 14

Since $\int_{3}^{\infty} \frac{dx}{x^2}$ is a convergent p-integral ( $p = 2 > 1$ ), by the Direct Comparison Test, our original integral $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ also converges!

STEP 15

(10) (a) (11) (c) (12) (c) (13) (c)

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Studdy Solution

STEP 1

STEP 2

1. Analyze the sequence2. Tackle the first integral3. Investigate the series4. Conquer the second integral

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

We know that ∫2∞dxx3/2\int_{2}^{\infty} \frac{dx}{x^{3/2}}∫2∞​x3/2dx​ **converges** because it's a p-integral with p=32>1p = \frac{3}{2} > 1p=23​>1.Therefore, our original integral also **converges**!

STEP 8

STEP 9

STEP 10

Since ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​ is a convergent p-series (p=2>1p = 2 > 1p=2>1), our original series also **converges**!

STEP 11

We're dealing with ∫3∞ln⁡2xx3dx\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx∫3∞​x3ln2x​dx.We know that for x>1x > 1x>1, 0<ln⁡x<x0 < \ln x < x0<lnx<x.Let's square this inequality: 0<ln⁡2x<x20 < \ln^2 x < x^20<ln2x<x2.

STEP 12

Now, divide by x3x^3x3: 0<ln⁡2xx3<x2x3=1x0 < \frac{\ln^2 x}{x^3} < \frac{x^2}{x^3} = \frac{1}{x}0<x3ln2x​<x3x2​=x1​.This doesn't help us much, since ∫3∞dxx\int_{3}^{\infty} \frac{dx}{x}∫3∞​xdx​ diverges.

STEP 13

Instead, we can use that for large xxx, ln⁡x\ln xlnx grows much slower than any positive power of xxx.So, for sufficiently large xxx, ln⁡2x<x\ln^2 x < xln2x<x.Therefore, ln⁡2xx3<xx3=1x2\frac{\ln^2 x}{x^3} < \frac{x}{x^3} = \frac{1}{x^2}x3ln2x​<x3x​=x21​.

STEP 14

Since ∫3∞dxx2\int_{3}^{\infty} \frac{dx}{x^2}∫3∞​x2dx​ is a convergent p-integral (p=2>1p = 2 > 1p=2>1), by the **Direct Comparison Test**, our original integral ∫3∞ln⁡2xx3dx\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx∫3∞​x3ln2x​dx also **converges**!

STEP 15

(10) (a) (11) (c) (12) (c) (13) (c)

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1. Analyze the sequence
2. Tackle the first integral
3. Investigate the series
4. Conquer the second integral

We know that $\int_{2}^{\infty} \frac{dx}{x^{3/2}}$ converges because it's a p-integral with $p = \frac{3}{2} > 1$ .
Therefore, our original integral also converges!

Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ ), our original series also converges!

We're dealing with $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ .
We know that for $x > 1$ , $0 < \ln x < x$ .
Let's square this inequality: $0 < \ln^2 x < x^2$ .

Now, divide by $x^3$ : $0 < \frac{\ln^2 x}{x^3} < \frac{x^2}{x^3} = \frac{1}{x}$ .
This doesn't help us much, since $\int_{3}^{\infty} \frac{dx}{x}$ diverges.

Instead, we can use that for large $x$ , $\ln x$ grows much slower than any positive power of $x$ .
So, for sufficiently large $x$ , $\ln^2 x < x$ .
Therefore, $\frac{\ln^2 x}{x^3} < \frac{x}{x^3} = \frac{1}{x^2}$ .

Since $\int_{3}^{\infty} \frac{dx}{x^2}$ is a convergent p-integral ( $p = 2 > 1$ ), by the Direct Comparison Test, our original integral $\int_{3}^{\infty} \frac{\ln^2 x}{x^3} dx$ also converges!