Solveeit Logo
Login

Question

Question: How do you use the integral test to determine if \(\sum {\dfrac{{\ln n}}{{{n^2}}}} \) from \(\left[ ...

How do you use the integral test to determine if ∑ln⁡nn2\sum {\dfrac{{\ln n}}{{{n^2}}}} from [1,∞)\left[ {1,\infty } \right) is convergent or divergent?

Explanation

Solution

Integral test is a method to check whether the given series converges or diverges. If the series converges we can find the summation of the series, and if the series diverges the summation of the series approaches infinity. To check for convergence or divergence using integral test we try to convert the given series into an integral of a continuous decreasing function and find its value.
i.e. ∑1∞an\sum\limits_1^\infty {{a_n}} is converted into ∫1∞f(x)dx\int\limits_1^\infty {f\left( x \right)dx} where ff is a continuous decreasing function in [1,∞)\left[ {1,\infty } \right).

Complete step by step solution:
We have been given a series ∑ln⁡nn2\sum {\dfrac{{\ln n}}{{{n^2}}}} and we have to use the integral test to determine whether the given series is convergent or divergent from [1,∞)\left[ {1,\infty } \right).
Integral test is given as,
For any continuous decreasing function ff in the interval [p,∞)\left[ {p,\infty } \right), the series is convergent if ∑p∞an<∫p∞f(x)dx\sum\limits_p^\infty {{a_n}} < \int\limits_p^\infty {f\left( x \right)dx} , and the series is divergent if ∫p∞f(x)dx<∑p∞an\int\limits_p^\infty {f\left( x \right)dx} < \sum\limits_p^\infty {{a_n}} .
In other words we can say that if we get a finite value of ∫p∞f(x)dx\int\limits_p^\infty {f\left( x \right)dx} , the given series is convergent. And in case we don’t get a finite value of ∫p∞f(x)dx\int\limits_p^\infty {f\left( x \right)dx} , the given series is divergent.
For the given series∑1∞ln⁡nn2\sum\limits_1^\infty {\dfrac{{\ln n}}{{{n^2}}}} , we can assume the function f(x)f\left( x \right) to be,
f(x)=ln⁡xx2f\left( x \right) = \dfrac{{\ln x}}{{{x^2}}}
Thus, we have to now evaluate the integral ∫1∞ln⁡xx2dx\int\limits_1^\infty {\dfrac{{\ln x}}{{{x^2}}}dx} .
We can use integration by parts to solve the above integral. Integration by parts is given as,
∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)dx} } }
Let u=ln⁡xu = \ln x and v=1x2v = \dfrac{1}{{{x^2}}}.
Then,
∫ln⁡xx2dx=ln⁡x∫1x2dx−∫(d(ln⁡x)dx∫1x2dx)dx\int {\dfrac{{\ln x}}{{{x^2}}}dx} = \ln x\int {\dfrac{1}{{{x^2}}}dx - \int {\left( {\dfrac{{d\left( {\ln x} \right)}}{{dx}}\int {\dfrac{1}{{{x^2}}}dx} } \right)} } dx
We can solve and find that ∫1x2dx=−1x+c\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c from the formula ∫xndx=x(n+1)(n+1)+c\int {{x^n}} dx = \dfrac{{{x^{\left( {n + 1} \right)}}}}{{\left( {n + 1} \right)}} + c.
And d(ln⁡x)dx=1x\dfrac{{d\left( {\ln x} \right)}}{{dx}} = \dfrac{1}{x} is known from special integral formula.
Thus, we can simplify our integral as,
=ln⁡x(−1x)−∫1x.(−1x)dx =−ln⁡xx+∫1x2dx =−ln⁡xx−1x+C =−ln⁡x−1x+C  = \ln x\left( { - \dfrac{1}{x}} \right) - \int {\dfrac{1}{x}.} \left( { - \dfrac{1}{x}} \right)dx \\\ = - \dfrac{{\ln x}}{x} + \int {\dfrac{1}{{{x^2}}}} dx \\\ = - \dfrac{{\ln x}}{x} - \dfrac{1}{x} + C \\\ = \dfrac{{ - \ln x - 1}}{x} + C \\\
Now we evaluate this integral in the limit,
∫1∞ln⁡xx2dx=[−ln⁡x−1x]1∞=[−ln⁡xx−1x]1∞ =[−ln⁡∞∞−1∞]−[−ln⁡11−11]  \int\limits_1^\infty {\dfrac{{\ln x}}{{{x^2}}}dx} = \left[ {\dfrac{{ - \ln x - 1}}{x}} \right]_1^\infty = \left[ { - \dfrac{{\ln x}}{x} - \dfrac{1}{x}} \right]_1^\infty \\\ = \left[ { - \dfrac{{\ln \infty }}{\infty } - \dfrac{1}{\infty }} \right] - \left[ { - \dfrac{{\ln 1}}{1} - \dfrac{1}{1}} \right] \\\
We can observe that lim⁡x→∞ln⁡xx=∞∞\mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln x}}{x} = \dfrac{\infty }{\infty }. So we will use L’Hospital Rule as follows,
lim⁡x→∞ln⁡xx=lim⁡x→∞d(ln⁡x)dxd(x)dx=lim⁡x→∞(1x)1=lim⁡x→∞1x=1∞=0\mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln x}}{x} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{{d\left( {\ln x} \right)}}{{dx}}}}{{\dfrac{{d\left( x \right)}}{{dx}}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {\dfrac{1}{x}} \right)}}{1} = \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = \dfrac{1}{\infty } = 0
Thus, we can simplify our integral as,
[−ln⁡∞∞−1∞]−[−ln⁡11−11]=[0−0]−[0−1]=1\left[ { - \dfrac{{\ln \infty }}{\infty } - \dfrac{1}{\infty }} \right] - \left[ { - \dfrac{{\ln 1}}{1} - \dfrac{1}{1}} \right] = \left[ {0 - 0} \right] - \left[ {0 - 1} \right] = 1
Thus, we get ∫1∞ln⁡xx2dx=1\int\limits_1^\infty {\dfrac{{\ln x}}{{{x^2}}}dx} = 1 which is a finite value.
Hence, the given series ∑ln⁡nn2\sum {\dfrac{{\ln n}}{{{n^2}}}} from [1,∞)\left[ {1,\infty } \right) is convergent.

Note: We used the integral test to check whether the given series is convergent or divergent. To check this we assumed a continuous function similar to the general term of the series and integrated upon this function within the given limits. As we got a finite value we said the given series is convergent. Once converting the given series into integral, we can calculate the integral using formula or tricks as required.