Estimating the sum ∑ k = 2 ∞ 1 k ln 2 ⁡ ( k )

Question

Estimating the sum       ∑          k      =      2              ∞            1          k              ln        2            ⁡      (      k      )

Hamnetmj · Accepted Answer

Answer:                              ∑                      k            =            2                                ∞                                    1                      k                          ln              2                        ⁡            k                                              =                  1                      2                          ln              2                        ⁡            2                          +                  1                      3                          ln              2                        ⁡            3                          +                  ∑                      k            =            4                                ∞                                    1                      k                          ln              2                        ⁡            k                                                            ≥                  1                      2                          ln              2                        ⁡            2                          +                  1                      3                          ln              2                        ⁡            3                          +                  ∫          4                      ∞                                                d            x                                x                          ln              2                        ⁡            x                                                            =                  1                      2                          ln              2                        ⁡            2                          +                  1                      3                          ln              2                        ⁡            3                          +                  1                      ln            ⁡            4                          &amp;gt;        2.038.

hogwartsxhoe5t · Accepted Answer

Using the Euler-Maclaurin Sum Formula, we get

\begin{aligned} \sum_{k = 2}^{n} \frac{1}{k \log (k)^{2}} \\ = C - \frac{1}{\log (n)} + \frac{1}{2 n \log (n)^{2}} - \frac{1}{12 n^{2}} (\frac{1}{\log (n)^{2}} + \frac{2}{\log (n)^{3}}) \\ + \frac{1}{360 n^{4}} (\frac{3}{\log (n)^{2}} + \frac{11}{\log (n)^{3}} + \frac{18}{\log (n)^{4}} + \frac{12}{\log (n)^{5}}) \\ - \frac{1}{15120 n^{6}} (\frac{60}{\log (n)^{2}} + \frac{274}{\log (n)^{3}} + \frac{675}{\log (n)^{4}} + \frac{1020}{\log (n)^{5}} + \frac{900}{\log (n)^{6}} + \frac{360}{\log (n)^{7}}) \\ + \frac{1}{50400 n^{8}} (\frac{210}{\log (n)^{2}} + \frac{1089}{\log (n)^{3}} + \frac{3283}{\log (n)^{4}} + \frac{6769}{\log (n)^{5}} + \frac{9800}{\log (n)^{6}} + \frac{9660}{\log (n)^{7}} + \frac{5880}{\log (n)^{8}} + \frac{1680}{\log (n)^{9}}) \\ - \frac{1}{1995840 n^{10}} (\frac{15120}{\log (n)^{2}} + \frac{85548}{\log (n)^{3}} + \frac{293175}{\log (n)^{4}} + \frac{723680}{\log (n)^{5}} + \frac{1346625}{\log (n)^{6}} + \frac{1898190}{\log (n)^{7}} + \frac{1984500}{\log (n)^{8}} + \frac{1461600}{\log (n)^{9}} + \frac{680400}{\log (n)^{10}} + \frac{151200}{\log (n)^{11}}) \\ (1) & + O (\frac{1}{n^{12} \log (n)^{2}}) \end{aligned}

Therefore, plugging

n = 1000

into (1), we get

\begin{aligned} \sum_{k = 2}^{\infty} \frac{1}{k \log (k)^{2}} & = C \\ (2) & = 2.109742801236891974479257197616551326 \end{aligned}

Estimating the sum sum_(k=2)^infty 1/(k ln^2(k))

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