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Calculus Level 5

∑ k = 0 ∞ ( 3 ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 3 ) 4 k + 3 − ln ⁡ ( 4 k + 4 ) 4 k + 4 − ln ⁡ ( 4 k + 5 ) 4 k + 5 ) = 3 ln ⁡ 2 2 − ln ⁡ 3 3 − ln ⁡ 4 4 − ln ⁡ 5 5 + 3 ln ⁡ 6 6 − ln ⁡ 7 7 − ln ⁡ 8 8 − ln ⁡ 9 9 + 3 ln ⁡ 10 10 − ⋯ \begin{aligned} & \sum_{k=0}^{\infty}\left(\frac{3\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right) \\ & =\frac{3\ln 2}2-\frac{\ln 3}3-\frac{\ln 4}4-\frac{\ln 5}5+\frac{3\ln 6}6-\frac{\ln 7}7-\frac{\ln 8}8-\frac{\ln 9}9+\frac{3\ln 10}{10}-\cdots \end{aligned}

Evaluate the sum above. Give your answer to 2 decimal places.

Notation: ln ⁡ ( ⋅ ) \ln (\cdot) denotes the natural logarithm.


The answer is 0.48.

Rajyawardhan Singh by Rajyawardhan Singh , 19, India

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1 solution

Thought it was insanely difficult during the contest because the sum doesn't converge absolutely and now I think I got it. Rip. :(

To avoid dealing with problems in absolute convergence, we deal with the n n -th partial sum. That is, ∑ k = 0 n ( 3 ⋅ ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 3 ) 4 k + 3 − ln ⁡ ( 4 k + 4 ) 4 k + 4 − ln ⁡ ( 4 k + 5 ) 4 k + 5 ) \sum_{k=0}^{n}\left(3\cdot\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right) = 3 ⋅ ln ⁡ 2 2 − ln ⁡ 3 3 − ln ⁡ 4 4 − ln ⁡ 5 5 + ⋯ + 3 ⋅ ln ⁡ ( 4 n + 2 ) 4 n + 2 − ln ⁡ ( 4 n + 3 ) 4 n + 3 − ln ⁡ ( 4 n + 4 ) 4 n + 4 − ln ⁡ ( 4 n + 5 ) 4 n + 5 . =3\cdot\frac{\ln 2}2-\frac{\ln 3}3-\frac{\ln 4}4-\frac{\ln 5}5+\cdots +3\cdot\frac{\ln(4n+2)}{4n+2}-\frac{\ln(4n+3)}{4n+3}-\frac{\ln(4n+4)}{4n+4}-\frac{\ln(4n+5)}{4n+5}. Because the sum is finite here, we have no issue of convergence and therefore can do the following conversion: ∑ k = 0 n ( 3 ⋅ ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 3 ) 4 k + 3 − ln ⁡ ( 4 k + 4 ) 4 k + 4 − ln ⁡ ( 4 k + 5 ) 4 k + 5 ) \sum_{k=0}^{n}\left(3\cdot\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right) = ∑ k = 0 n ( ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 3 ) 4 k + 3 + ln ⁡ ( 4 k + 4 ) 4 k + 4 − ln ⁡ ( 4 k + 5 ) 4 k + 5 ) + ∑ k = 0 n 2 ( ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 4 ) 4 k + 4 ) =\sum_{k=0}^{n}\left(\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}+\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right) +\sum_{k=0}^{n}2\left(\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+4)}{4k+4}\right) Also notice that ∑ k = 0 n 2 ( ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 4 ) 4 k + 4 ) = ∑ k = 0 n ( ln ⁡ 2 + ln ⁡ ( 2 k + 1 ) 2 k + 1 − ln ⁡ 2 + ln ⁡ ( 2 k + 2 ) 2 k + 2 ) \sum_{k=0}^{n}2\left(\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+4)}{4k+4}\right) =\sum_{k=0}^{n}\left(\frac{\ln 2 + \ln(2k+1)}{2k+1}-\frac{\ln 2 + \ln(2k+2)}{2k+2}\right) So we have ∑ k = 0 n ( ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 3 ) 4 k + 3 + ln ⁡ ( 4 k + 4 ) 4 k + 4 − ln ⁡ ( 4 k + 5 ) 4 k + 5 ) + ∑ k = 0 n 2 ( ln ⁡ ( 4 k + 2 ) 4 k + 2 − ln ⁡ ( 4 k + 4 ) 4 k + 4 ) \sum_{k=0}^{n}\left(\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}+\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right) +\sum_{k=0}^{n}2\left(\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+4)}{4k+4}\right) = ∑ k = 0 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) + ∑ k = 0 n ( ln ⁡ 2 + ln ⁡ ( 2 k + 1 ) 2 k + 1 − ln ⁡ 2 + ln ⁡ ( 2 k + 2 ) 2 k + 2 ) =\sum_{k=0}^{2n+1}\left(\frac{\ln(2k+2)}{2k+2}-\frac{\ln(2k+3)}{2k+3}\right) +\sum_{k=0}^{n}\left(\frac{\ln 2 + \ln(2k+1)}{2k+1}-\frac{\ln 2 + \ln(2k+2)}{2k+2}\right) = ln ⁡ 2 ∑ k = 0 n ( 1 2 k + 1 − 1 2 k + 2 ) − ln ⁡ ( 2 n + 2 ) 2 n + 2 + ∑ k = n + 1 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) =\ln 2\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)-\frac {\ln (2n+2)}{2n+2} +\sum_{k=n+1}^{2n+1}\left(\frac{\ln(2k+2)}{2k+2}-\frac{\ln(2k+3)}{2k+3}\right) Now, notice that ln ⁡ x x \frac {\ln x}{x} is a decreasing sequence with limit 0 0 as x → ∞ x\to\infty . Thus ∑ k = 0 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) \sum_{k=0}^{2n+1}\left(\frac{\ln(2k+2)}{2k+2}-\frac{\ln(2k+3)}{2k+3}\right) is an alternating sum hence converges), which means that ∑ k = n + 1 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) → 0 \sum_{k=n+1}^{2n+1}\left(\frac{\ln(2k+2)}{2k+2}-\frac{\ln(2k+3)}{2k+3}\right)\to 0 as n → 0 n\to 0 . It is also well known that ∑ k = 0 n ( 1 2 k + 1 − 1 2 k + 2 ) → ln ⁡ 2 \sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\to\ln 2 as n → ∞ n\to\infty . Therefore lim ⁡ n → ∞ ln ⁡ 2 ∑ k = 0 n ( 1 2 k + 1 − 1 2 k + 2 ) − ln ⁡ ( 2 n + 2 ) 2 n + 2 + ∑ k = n + 1 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) = ( ln ⁡ 2 ) 2 + 0 + 0 = ( ln ⁡ 2 ) 2 \lim_{n\to\infty}\ln 2\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)-\frac {\ln (2n+2)}{2n+2} +\sum_{k=n+1}^{2n+1}\left(\frac{\ln(2k+2)}{2k+2}-\frac{\ln(2k+3)}{2k+3}\right) =(\ln 2)^2+0+0=(\ln 2)^2 Giving ( 0.693 ) 2 (0.693)^2 = taking 0.48 0.48

6 Helpful 1 Interesting 1 Brilliant 0 Confused

Great solution, but I think there’s a typo after ∑ k = n + 1 2 n + 1 ( ln ⁡ ( 2 k + 2 ) 2 k + 2 − ln ⁡ ( 2 k + 3 ) 2 k + 3 ) → 0 \sum_{k=n+1}^{2n+1} \left(\dfrac{\ln{\left( 2k+2 \right)}}{2k+2}-\dfrac{\ln{\left( 2k+3 \right)}}{2k+3} \right) \rightarrow 0 .

This sum approaches zero as n → ∞ n \to \boxed{\infty} .

Akeel Howell - 3 years, 2 months ago

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