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Question: How do you test the series \(\dfrac{\ln k}{{{k}^{2}}}\) for convergence?...

How do you test the series ln⁡kk2\dfrac{\ln k}{{{k}^{2}}} for convergence?

Explanation

Solution

Now to solve this problem we will use an integral test for convergence. First we will check if the initial conditions are satisfied by the functions. Hence if the function is positive, continuous and decreasing in the interval [1,∞]\left[ 1,\infty \right] we can proceed with the integral test. Hence we will find if the integral ∫1∞ln⁡xx2\int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}} converges. The nature of the series will be the same as the nature of the integral according to the Integral test.

Complete step-by-step solution:
Consider the given series ∑ln⁡kk2\sum{\dfrac{\ln k}{{{k}^{2}}}} .
To check the convergence of the series we will use Integral test.
Hence, the function f(x)=ln⁡xx2f\left( x \right)=\dfrac{\ln x}{{{x}^{2}}}
Now first let us check if all the conditions of the Integral test theorem are satisfied.
To use the integral test the function must be positive continuous and decreasing in the interval [1,∞]\left[ 1,\infty \right] where f(n)=anf\left( n \right)={{a}_{n}} where an{{a}_{n}} is nth{{n}^{th}} term of the sequence.
Now we know that ln⁡x>0\ln x>0 and x2>0{{x}^{2}}>0 for all value of x and hence the function is also positive for x∈[1,∞]x\in \left[ 1,\infty \right]
Hence f(x)>0f\left( x \right)>0 for x∈[1,∞]x\in \left[ 1,\infty \right]
Now let us check the difference in the given function.
f(x)=ln⁡xx2f\left( x \right)=\dfrac{\ln x}{{{x}^{2}}}
Now we know that if f(x)=pq⇒f′(x)=qp′−pq′q2f\left( x \right)=\dfrac{p}{q}\Rightarrow f'\left( x \right)=\dfrac{qp'-pq'}{{{q}^{2}}}
Hence differentiating the function ln⁡xx2\dfrac{\ln x}{{{x}^{2}}} with the above formula we get, x21x−2x(ln⁡x)x4=(1−2ln⁡x)x3\dfrac{{{x}^{2}}\dfrac{1}{x}-2x\left( \ln x \right)}{{{x}^{4}}}=\dfrac{\left( 1-2\ln x \right)}{{{x}^{3}}}
Now we know that in the interval [1,∞]\left[ 1,\infty \right] x3>0{{x}^{3}}>0 and 1−2ln⁡x<01-2\ln x<0 as ln⁡x>1\ln x>1 for x>1x>1
Hence we can say that the function 1−2ln⁡xx3<0\dfrac{1-2\ln x}{{{x}^{3}}}<0 in the interval [1,∞]\left[ 1,\infty \right]
Hence we have f′(x)<0f'\left( x \right)<0 in the interval [1,∞]\left[ 1,\infty \right]
Now let us check lim⁡x→∞ f(x)\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)
⇒lim⁡x→∞ f(x)=lim⁡x→∞ ln⁡xx2\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\ln x}{{{x}^{2}}}
Using L-hospital rule we get that the limit is equal to lim⁡x→∞ 1x2x=lim⁡x→∞ 12x2=0\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{1}{x}}{2x}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2{{x}^{2}}}=0
Hence function is monotone decreasing in the interval [1,∞]\left[ 1,\infty \right]
Now we can see the conditions for the integral test are satisfied.
Now we can say that if ∫1∞ln⁡xx2\int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}} converges then the series ∑ln⁡xx2\sum{\dfrac{\ln x}{{{x}^{2}}}} converges.
Now consider the integral ∫1∞ln⁡xx2\int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}
Now we know that ∫f.g=f∫g−∫(f′∫g)\int{f.g}=f\int{g}-\int{\left( f'\int{g} \right)} hence using this we get,
⇒∫1∞ln⁡xx2=[ln⁡x(−1x)−∫(1x)(−1x)]1∞ ⇒∫1∞ln⁡xx2=[−ln⁡xx+∫1x2]1∞ ⇒∫1∞ln⁡xx2=[−ln⁡xx−1x]1∞ ⇒∫1∞ln⁡xx2=−[ln⁡x+1x]1∞ ⇒∫1∞ln⁡xx2=−[lim⁡x→∞ (ln⁡x+1x)−ln⁡1+11] ⇒∫1∞ln⁡xx2=−[0−0+11]=1 \begin{aligned} & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=\left[ \ln x\left( -\dfrac{1}{x} \right)-\int{\left( \dfrac{1}{x} \right)\left( \dfrac{-1}{x} \right)} \right]_{1}^{\infty } \\\ & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=\left[ -\dfrac{\ln x}{x}+\int{\dfrac{1}{{{x}^{2}}}} \right]_{1}^{\infty } \\\ & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=\left[ -\dfrac{\ln x}{x}-\dfrac{1}{x} \right]_{1}^{\infty } \\\ & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=-\left[ \dfrac{\ln x+1}{x} \right]_{1}^{\infty } \\\ & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=-\left[ \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\ln x+1}{x} \right)-\dfrac{\ln 1+1}{1} \right] \\\ & \Rightarrow \int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}}=-\left[ 0-\dfrac{0+1}{1} \right]=1 \\\ \end{aligned}
Hence we get the integral ∫1∞ln⁡xx2\int_{1}^{\infty }{\dfrac{\ln x}{{{x}^{2}}}} converges.
Hence by Integral test we have the series ∑ln⁡xx2\sum{\dfrac{\ln x}{{{x}^{2}}}} also converges.

Note: Note that since the function obtained can be easily integrated we go for integral tests to check the convergence of the series instead of comparison tests. Also while using Integral test always check if the function satisfies all the initial conditions and then proceed with the test.