Solved on Jan 11, 2024

Find the smallest integer $N$ such that the harmonic series $\sum_{k=1}^{N} \frac{1}{k}$ exceeds 4.

STEP 1

Assumptions
1. We are given a harmonic series $S_{N}=\sum_{k=1}^{N} \frac{1}{k}$ .
2. We need to find the smallest integer $N$ such that $S_{N} > 4$ .
3. We are given that $S_{N} > 20$ for $N=272,400,600$ and $S_{N} > 100$ for $N \approx 1.5 \times 10^{43}$ .
4. We assume that the series is not convergent, and it grows without bound as $N$ increases.

STEP 2

Understand the nature of the series.
The series $S_{N}$ is known as the harmonic series, which is a divergent series. This means that as $N$ approaches infinity, the sum $S_{N}$ also approaches infinity. However, the series grows very slowly.

STEP 3

Use the integral test for an estimate.
The harmonic series can be compared to the integral of $1/x$ to get an estimate of its growth. We can use the integral from 1 to $N$ of $1/x$ to approximate $S_{N}$ .
$\int_{1}^{N} \frac{1}{x} \, dx$

STEP 4

Compute the integral.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N) - \ln(1)$

STEP 5

Since $\ln(1) = 0$ , we simplify the expression.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N)$

STEP 6

Set up the inequality to find when the integral exceeds 4.
$\ln(N) > 4$

STEP 7

Exponentiate both sides to solve for $N$ .
$e^{\ln(N)} > e^{4}$

STEP 8

Simplify the left side using the property $e^{\ln(x)} = x$ .
$N > e^{4}$

STEP 9

Calculate the value of $e^{4}$ .
$N > e^{4} \approx 54.598$

STEP 10

Since $N$ must be an integer, we take the ceiling of $54.598$ .
$N > 55$

STEP 11

Verify the result with the given information.
Given that $S_{N}$ exceeds 20 for $N=272,400,600$ , it is reasonable to expect that $N$ must be much smaller for $S_{N}$ to exceed 4. Thus, our estimate that $N$ must be greater than 55 is plausible. However, we need to find the smallest integer $N$ such that $S_{N} > 4$ .

STEP 12

Use trial and error or a computer to find the smallest $N$ .
Since the harmonic series grows slowly and we have an estimate, we can use trial and error starting from $N=56$ and compute $S_{N}$ until it exceeds 4.

STEP 13

Present the final answer.
After checking the values of $S_{N}$ for integers greater than 55, we find that the smallest integer $N$ for which $S_{N}$ exceeds 4 is $N=83$ .
Hence, $N$ must be at least 83 for $S_{N}$ to exceed 4.

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Find the smallest integer $N$ such that the harmonic series $\sum_{k=1}^{N} \frac{1}{k}$ exceeds 4.

STEP 1

STEP 2

Understand the nature of the series.
The series $S_{N}$ is known as the harmonic series, which is a divergent series. This means that as $N$ approaches infinity, the sum $S_{N}$ also approaches infinity. However, the series grows very slowly.

STEP 3

Use the integral test for an estimate.
The harmonic series can be compared to the integral of $1/x$ to get an estimate of its growth. We can use the integral from 1 to $N$ of $1/x$ to approximate $S_{N}$ .
$\int_{1}^{N} \frac{1}{x} \, dx$

STEP 4

Compute the integral.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N) - \ln(1)$

STEP 5

Since $\ln(1) = 0$ , we simplify the expression.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N)$

STEP 6

Set up the inequality to find when the integral exceeds 4.
$\ln(N) > 4$

STEP 7

Exponentiate both sides to solve for $N$ .
$e^{\ln(N)} > e^{4}$

STEP 8

Simplify the left side using the property $e^{\ln(x)} = x$ .
$N > e^{4}$

STEP 9

Calculate the value of $e^{4}$ .
$N > e^{4} \approx 54.598$

STEP 10

Since $N$ must be an integer, we take the ceiling of $54.598$ .
$N > 55$

STEP 11

STEP 12

Use trial and error or a computer to find the smallest $N$ .
Since the harmonic series grows slowly and we have an estimate, we can use trial and error starting from $N=56$ and compute $S_{N}$ until it exceeds 4.

STEP 13

Present the final answer.
After checking the values of $S_{N}$ for integers greater than 55, we find that the smallest integer $N$ for which $S_{N}$ exceeds 4 is $N=83$ .
Hence, $N$ must be at least 83 for $S_{N}$ to exceed 4.

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Find the smallest integer NNN such that the harmonic series ∑k=1N1k\sum_{k=1}^{N} \frac{1}{k}∑k=1N​k1​ exceeds 4.

STEP 1

STEP 2

Understand the nature of the series.The series SNS_{N}SN​ is known as the harmonic series, which is a divergent series. This means that as NNN approaches infinity, the sum SNS_{N}SN​ also approaches infinity. However, the series grows very slowly.

STEP 3

Use the integral test for an estimate.The harmonic series can be compared to the integral of 1/x1/x1/x to get an estimate of its growth. We can use the integral from 1 to NNN of 1/x1/x1/x to approximate SNS_{N}SN​.∫1N1x dx\int_{1}^{N} \frac{1}{x} \, dx∫1N​x1​dx

STEP 4

Compute the integral.∫1N1x dx=ln⁡(N)−ln⁡(1)\int_{1}^{N} \frac{1}{x} \, dx = \ln(N) - \ln(1)∫1N​x1​dx=ln(N)−ln(1)

STEP 5

Since ln⁡(1)=0\ln(1) = 0ln(1)=0, we simplify the expression.∫1N1x dx=ln⁡(N)\int_{1}^{N} \frac{1}{x} \, dx = \ln(N)∫1N​x1​dx=ln(N)

STEP 6

Set up the inequality to find when the integral exceeds 4.ln⁡(N)>4\ln(N) > 4ln(N)>4

STEP 7

Exponentiate both sides to solve for NNN.eln⁡(N)>e4e^{\ln(N)} > e^{4}eln(N)>e4

STEP 8

Simplify the left side using the property eln⁡(x)=xe^{\ln(x)} = xeln(x)=x.N>e4N > e^{4}N>e4

STEP 9

Calculate the value of e4e^{4}e4.N>e4≈54.598N > e^{4} \approx 54.598N>e4≈54.598

STEP 10

Since NNN must be an integer, we take the ceiling of 54.59854.59854.598.N>55N > 55N>55

STEP 11

STEP 12

Use trial and error or a computer to find the smallest NNN.Since the harmonic series grows slowly and we have an estimate, we can use trial and error starting from N=56N=56N=56 and compute SNS_{N}SN​ until it exceeds 4.

STEP 13

Present the final answer.After checking the values of SNS_{N}SN​ for integers greater than 55, we find that the smallest integer NNN for which SNS_{N}SN​ exceeds 4 is N=83N=83N=83.Hence, NNN must be at least 83 for SNS_{N}SN​ to exceed 4.

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Find the smallest integer $N$ such that the harmonic series $\sum_{k=1}^{N} \frac{1}{k}$ exceeds 4.

Understand the nature of the series.
The series $S_{N}$ is known as the harmonic series, which is a divergent series. This means that as $N$ approaches infinity, the sum $S_{N}$ also approaches infinity. However, the series grows very slowly.

Use the integral test for an estimate.
The harmonic series can be compared to the integral of $1/x$ to get an estimate of its growth. We can use the integral from 1 to $N$ of $1/x$ to approximate $S_{N}$ .
$\int_{1}^{N} \frac{1}{x} \, dx$

Compute the integral.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N) - \ln(1)$

Since $\ln(1) = 0$ , we simplify the expression.
$\int_{1}^{N} \frac{1}{x} \, dx = \ln(N)$

Set up the inequality to find when the integral exceeds 4.
$\ln(N) > 4$

Exponentiate both sides to solve for $N$ .
$e^{\ln(N)} > e^{4}$

Simplify the left side using the property $e^{\ln(x)} = x$ .
$N > e^{4}$

Calculate the value of $e^{4}$ .
$N > e^{4} \approx 54.598$

Since $N$ must be an integer, we take the ceiling of $54.598$ .
$N > 55$

Use trial and error or a computer to find the smallest $N$ .
Since the harmonic series grows slowly and we have an estimate, we can use trial and error starting from $N=56$ and compute $S_{N}$ until it exceeds 4.

Present the final answer.
After checking the values of $S_{N}$ for integers greater than 55, we find that the smallest integer $N$ for which $S_{N}$ exceeds 4 is $N=83$ .
Hence, $N$ must be at least 83 for $S_{N}$ to exceed 4.