Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. ln⁡22+ln⁡33+ln⁡44+ln⁡55+ln⁡66+...

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Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.  ln⁡22+ln⁡33+ln⁡44+ln⁡55+ln⁡66+...

rogreenhoxa8 · Accepted Answer

We have the given series is,ln⁡22+ln⁡33+ln⁡44+ln⁡55+ln⁡66+...=∑n=2∞ln⁡(n)nBefore getting into the integral test, we must assure two things first : for the integral step to apply to∑n=N∞f(n) we must have an&amp;gt;0 for the given interval and f(n) must be decresing on the same interval.Both of these are true sinceln⁡(n)&amp;gt;0 and n&amp;gt;0 on n∈[2,∞) so ln⁡(n)n&amp;gt;0 on the same interval.Furthermore, ln⁡(n) grows slower than the n so we see that ln⁡(n)n is decresing because n overpowers ln⁡(n) in the numerator.We can also sjow this by taking derivative of ln⁡(n)n and showing it&#039;s away negative on n∈[2,∞).So wee see the integral test applies.The integral test states that if the two if two condition are met,then for ∑n=N∞f(n), evaluate the improper integral ∫N∞f(x)dxIf the integral converges to a real,finite value,then the series converges.If the integral diverges, then the series does too.So, we take the integral ∫2∞ln⁡(x)xdxNow by taking the limit as it goes to infinity∫2∞ln⁡(x)xdx=limb→∞∫2∞ln⁡(x)xdxletting u=ln⁡(x)so du=1xdx=limb→∞∫ln⁡(2)ln⁡(b)udu=limb→∞[12u2]ln⁡(2)ln⁡(b)=limb→∞12ln2⁡(b)−12ln2⁡(2)as b→∞, we see that ln⁡(b)→∞, so,=∞hence, the integral diverges.Thus, we see that∑n=2∞ln⁡(n)n diverges as well

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. frac{ln2}{2}+frac{ln3}{3}+frac{ln4}{4}+frac{ln5}{5}+frac{ln6}{6}+...

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