Euler-Maclaurin summation

[24/38%]
Show
LaTeX
Problem 5251 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove, for m>1m>1, the Euler--Maclaurin formula:

∑n=a∞1nm=1(m−1)am−1+121am+B12!mam+1−B24!m(m+1)(m+2)am+3+…+(−1)p−1Bp(2p)!m(m+1)…(m+2p−2)am+2p−1θ;0<θ<1, \sum _{n=a}^{\infty } \frac{1}{n^{m}} = \frac{1}{(m-1)a^{m-1}} + \frac{1}{2}\frac{1}{a^{m}} + \frac{B_{1}}{2!}\frac{m}{a^{m+1}} - \frac{B_{2}}{4!}\frac{m(m+1)(m+2)}{a^{m+3}} + \ldots + (-1)^{p-1}\frac{B_{p}}{(2p)!}\frac{m(m+1)\ldots (m+2p-2)}{a^{m+2p-1}}\theta ; \quad 0 < \theta < 1,

where B1,B2,B3,…B_{1}, B_{2}, B_{3}, \ldots are the Bernoulli numbers.

?
Problem 5252 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Setting σm=∑n=10∞1nm\sigma_{m} = \sum_{n=10}^{\infty } \dfrac {1}{n^{m}}, prove that the values of the first few σm\sigma_{m}, expressed in units of the 15th decimal digit, are as follows:

?
Problem 5253 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove the equality ∑m=2∞σm=19\displaystyle \sum_{m=2}^{\infty }\sigma_{m} = \dfrac {1}{9} and, using it, check the precision of the values of σm\sigma_{m} given in the preceding problem (problem 5252).

?
Problem 5254 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the convergence of the series ∑n=1∞(1n−ln⁡n+1n)\displaystyle \sum_{n=1}^{\infty }\left(\frac{1}{n} - \ln \frac{n+1}{n}\right), the Euler--Maclaurin formula, and the fact that

∑n=m∞un=lim⁡p→∞∑n=mm+pun, \sum _{n=m}^{\infty } u_{n} = \lim _{p\to \infty }\sum _{n=m}^{m+p} u_{n},

prove the equality:

1+12+13+…+1m=C+ln⁡m+12m−B12⋅1m2+…+(−1)p−1Bp−12p−21m2p−2+(−1)pBp2p1m2p⋅θ;0<θ<1. 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{m} = C + \ln m + \frac{1}{2m} - \frac{B_{1}}{2}\cdot \frac{1}{m^{2}} + \ldots + (-1)^{p-1}\frac{B_{p-1}}{2p-2}\frac{1}{m^{2p-2}} + (-1)^{p}\frac{B_{p}}{2p}\frac{1}{m^{2p}}\cdot \theta ; \quad 0 < \theta < 1.
?
Problem 5255 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Setting m=3m=3, p=9p=9 in the preceding result (problem 5254), compute the constant CC (Euler's constant).

?
Problem 5256 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Compute the sum ∑n=110001n\displaystyle \sum_{n=1}^{1000} \frac{1}{n} to within 10−710^{-7}.

?
Problem 5257 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the Euler--Maclaurin formula, prove the equality:

∑n=1m1n=A+2m+12m−B12! 2mm+1⋅3⋅54! 23B2m3m−…+(−1)p⋅1⋅3…(4p−3)Bp(2p)! 22p−1m2p−12+(−1)p+1⋅1⋅3…(4p−1)(2p+2)! 22p+1⋅Bp+1m2p+32⋅θ;0<θ<1. \sum _{n=1}^{m} \frac{1}{\sqrt{n}} = A + 2\sqrt{m} + \frac{1}{2\sqrt{m}} - \frac{B_{1}}{2!\, 2m\sqrt{m}} + \frac{1\cdot 3\cdot 5}{4!\, 2^{3}}\frac{B_{2}}{m^{3}\sqrt{m}} - \ldots + \frac{(-1)^{p}\cdot 1\cdot 3\ldots (4p-3)B_{p}}{(2p)!\, 2^{2p-1}m^{2p-\frac12}} + \frac{(-1)^{p+1}\cdot 1\cdot 3\ldots (4p-1)}{(2p+2)!\, 2^{2p+1}}\cdot \frac{B_{p+1}}{m^{2p+\frac32}}\cdot \theta ; \quad 0<\theta <1.
?
Problem 5258 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the Euler--Maclaurin formula, compute the sums of the series ∑n=1∞1n3\displaystyle \sum_{n=1}^{\infty }\frac{1}{n^{3}} and ∑n=1∞ln⁡nn2\displaystyle \sum_{n=1}^{\infty }\frac{\ln n}{n^{2}} to within 10−710^{-7}.

?
Problem 5259 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the integrals ∫0∞e−mx dx=1m\displaystyle \int_{0}^{\infty } e^{-mx}\, dx = \dfrac {1}{m} and ∫0∞e−mxsin⁡nx dx=nm2+n2\displaystyle \int_{0}^{\infty } e^{-mx}\sin nx\, dx = \dfrac {n}{m^{2}+n^{2}} (m>0m>0), prove the equality:

1n+1−1n+2+1n+3−…=∫0∞e−nxex+1 dx. \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \ldots = \int _{0}^{\infty } \frac{e^{-nx}}{e^{x}+1}\, dx.
?
Problem 5260 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the integrals ∫0∞e−mx dx=1m\displaystyle \int_{0}^{\infty } e^{-mx}\, dx = \dfrac {1}{m} and ∫0∞e−mxsin⁡nx dx=nm2+n2\displaystyle \int_{0}^{\infty } e^{-mx}\sin nx\, dx = \dfrac {n}{m^{2}+n^{2}} (m>0m>0), prove the equality:

12n+1−12n+3+12n+5−…=∫0∞e−2nxex+e−x dx. \frac{1}{2n+1} - \frac{1}{2n+3} + \frac{1}{2n+5} - \ldots = \int _{0}^{\infty } \frac{e^{-2nx}}{e^{x}+e^{-x}}\, dx.
?
Problem 5261 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the integrals ∫0∞e−mx dx=1m\displaystyle \int_{0}^{\infty } e^{-mx}\, dx = \dfrac {1}{m} and ∫0∞e−mxsin⁡nx dx=nm2+n2\displaystyle \int_{0}^{\infty } e^{-mx}\sin nx\, dx = \dfrac {n}{m^{2}+n^{2}} (m>0m>0), prove the equality:

∑ν=0∞(−1)νnn2+(n+ν+1)2=∫0∞e−nxsin⁡nxex+1 dx. \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }n}{n^{2}+(n+\nu +1)^{2}} = \int _{0}^{\infty } \frac{e^{-nx}\sin nx}{e^{x}+1}\, dx.
?
Problem 5262 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the series expansions of the fractions 1ex+1\frac{1}{e^{x}+1} and the preceding results, prove the equality:

ln⁡2=∑ν=1n(−1)ν−1ν+(−1)n2n+12∑μ=1m−1(−1)n+μ(22μ−1)Bμn2μμ+θ⋅(−1)n+m(22m−1)⋅Bmn2m⋅m. \ln 2 = \sum _{\nu =1}^{n} \frac{(-1)^{\nu -1}}{\nu } + \frac{(-1)^{n}}{2n} + \frac{1}{2}\sum _{\mu =1}^{m-1}\frac{(-1)^{n+\mu }(2^{2\mu }-1)B_{\mu }}{n^{2\mu }\mu } + \theta \cdot (-1)^{n+m}\frac{(2^{2m}-1)\cdot B_{m}}{n^{2m}\cdot m}.
?
Remark.
?

For any real xx the following formulas hold:

1ex+1=12+∑ν=1n−1(−1)ν(22ν−1)Bν(2ν)!x2ν−1+θ⋅(−1)n⋅(22n−1)(2n)!Bn⋅x2n−1. \frac{1}{e^{x}+1} = \frac{1}{2} + \sum _{\nu =1}^{n-1}(-1)^{\nu }\frac{(2^{2\nu }-1)B_{\nu }}{(2\nu )!}x^{2\nu -1} + \theta \cdot (-1)^{n}\cdot \frac{(2^{2n}-1)}{(2n)!}B_{n}\cdot x^{2n-1}. 1ex−1=1x−12+∑ν=1n−1(−1)ν−1Bν(2ν)!x2ν−1+θ⋅(−1)n−1Bn(2n)!x2n−1. \frac{1}{e^{x}-1} = \frac{1}{x} - \frac{1}{2} + \sum _{\nu =1}^{n-1}\frac{(-1)^{\nu -1}B_{\nu }}{(2\nu )!}x^{2\nu -1} + \theta \cdot (-1)^{n-1}\frac{B_{n}}{(2n)!}x^{2n-1}. 1ex+e−x=12∑ν=0n−1(−1)νEν(2ν)!x2ν+θ2⋅(−1)nEn(2n)!x2n. \frac{1}{e^{x}+e^{-x}} = \frac{1}{2}\sum _{\nu =0}^{n-1}\frac{(-1)^{\nu }E_{\nu }}{(2\nu )!}x^{2\nu } + \frac{\theta }{2}\cdot \frac{(-1)^{n}E_{n}}{(2n)!}x^{2n}.

Here BnB_{n} and EnE_{n} are the Bernoulli and Euler numbers, 0<θ<10 < \theta < 1.

Problem 5263 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using the series expansion of the fraction 1ex+e−x\frac{1}{e^{x}+e^{-x}} (see the remark to problem 5262) and the preceding results, prove the equality:

π4=∑ν=1n(−1)ν−12ν−1+n∑μ=1m(−1)n+μ−1Eμ−1(2n)2μ+nθ⋅(−1)n+m⋅Em(2n)2m+2. \frac{\pi }{4} = \sum _{\nu =1}^{n}\frac{(-1)^{\nu -1}}{2\nu -1} + n\sum _{\mu =1}^{m}\frac{(-1)^{n+\mu -1}E_{\mu -1}}{(2n)^{2\mu }} + \frac{n\theta \cdot (-1)^{n+m}\cdot E_{m}}{(2n)^{2m+2}}.
?
Problem 5264 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Using this and the preceding results (see the remark to problem 5262), prove the equality:

π2=∑ν=1n2nn2+ν2−2πe2πn−1+12n+∑μ=1m(−1)μ−1B2μ−1(2n2)2μ−112μ−1+θ(−1)m(2n2)2m+1B2m+12m+1. \frac{\pi }{2} = \sum _{\nu =1}^{n}\frac{2n}{n^{2}+\nu ^{2}} - \frac{2\pi }{e^{2\pi n}-1} + \frac{1}{2n} + \sum _{\mu =1}^{m}\frac{(-1)^{\mu -1}B_{2\mu -1}}{(2n^{2})^{2\mu -1}}\frac{1}{2\mu -1} + \frac{\theta (-1)^{m}}{(2n^{2})^{2m+1}}\frac{B_{2m+1}}{2m+1}.

Here 0<θ<10 < \theta < 1. The last formula was given by Euler without an indication of the remainder term.

?
Problem 5265 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove the formula:

∑ν=1nln⁡νν=C+12(ln⁡n)2+ln⁡n2n+1−ln⁡n12n2+θ6ln⁡n−11720n4, \sum _{\nu =1}^{n} \frac{\ln \nu }{\nu } = C + \frac{1}{2}(\ln n)^{2} + \frac{\ln n}{2n} + \frac{1-\ln n}{12n^{2}} + \theta \frac{6\ln n - 11}{720 n^{4}},

where C=−0,072815…C = -0{,}072815\ldots; 0<θ<10 < \theta < 1.

?
Problem 5266 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove the equality ∑ν=1∞ln⁡νν2=0,9375483…\displaystyle \sum_{\nu =1}^{\infty } \frac{\ln \nu }{\nu^{2}} = 0{,}9375483\ldots

?
Problem 5267 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove the equality ∑ν=1∞(−1)νln⁡νν2=0,1013166…\displaystyle \sum_{\nu =1}^{\infty } \frac{(-1)^{\nu }\ln \nu }{\nu^{2}} = 0{,}1013166\ldots

?
Problem 5268 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 13.3 "The Euler--Maclaurin Formula and Similar Methods"

Prove the equality ∑ν=1∞(−1)ν−1ν=0,604898…\displaystyle \sum_{\nu =1}^{\infty } \frac{(-1)^{\nu -1}}{\sqrt{\nu }} = 0{,}604898\ldots

?
Problem II.1.10 from Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. "Избранные задачи по вещественному анализу" [МГЛ+04]More problems in Section II.1 "Computing Limits"

Let f∈C1([0,1])f \in C^{1}(\left[0,1\right]). Prove that

∑1⩽k⩽nf(kn)−n∫01f(x)dx⟶n→∞f(1)−f(0)2 \sum _{1 \leqslant k \leqslant n} f\left(\frac{k}{n}\right)-n \int _{0}^{1} f(x) d x \underset {n \rightarrow \infty }{\longrightarrow } \frac{f(1)-f(0)}{2}
?
Problem II.2.11 from Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. "Избранные задачи по вещественному анализу" [МГЛ+04]More problems in Section II.2 "Averaging of Sequences"

Prove that as n→+∞n \rightarrow +\infty

?
(a)

∑1⩽k⩽nkp=np+1p+1+αp+o(1)\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\alpha_{p}+o(1) (p∈(−1,0]);(p \in (-1,0]) ;

(b)

∑1⩽k⩽nkp=np+1p+1+np2+βp+o(1)\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\frac{n^{p}}{2}+\beta_{p}+o(1) (p∈(0,1]);(p \in (0,1]) ;

(c)

∑1⩽k⩽nkp=np+1p+1+np2+p12np−1+γp+o(1)\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\frac{n^{p}}{2}+\frac{p}{12} n^{p-1}+\gamma_{p}+o(1) (p∈(1,2])(p \in (1,2]).

Problem II.2.12 from Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. "Избранные задачи по вещественному анализу" [МГЛ+04]More problems in Section II.2 "Averaging of Sequences"

Prove that as n→+∞n \rightarrow +\infty

?
(a)

∑1<k⩽n1kln⁡pk=1p+1ln⁡p+1n+Cp+o(1)(p>−1)\sum_{1<k \leqslant n} \frac{1}{k} \ln^{p} k=\frac{1}{p+1} \ln^{p+1} n+C_{p}+o(1) \quad (p>-1);

(b)

∑1<k⩽n1kln⁡k=ln⁡(ln⁡n)+C+o(1)\sum_{1<k \leqslant n} \frac{1}{k \ln k}=\ln (\ln n)+C+o(1);

(c)

∑1⩽k⩽n1k!k=eln⁡n+C′+o(1)\sum_{1 \leqslant k \leqslant n} \frac{1}{\sqrt[k]{k!}}=e \ln n+C^{\prime }+o(1).

Problem II.2.13 from Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. "Избранные задачи по вещественному анализу" [МГЛ+04]More problems in Section II.2 "Averaging of Sequences"

Prove that

∑1⩽k⩽nkln⁡k=(n2+n+16)ln⁡n2−14n2+C1+o(1),∑1⩽k⩽nln⁡k!=(n2+2n+12)ln⁡n2−3n24+(ln⁡2π2−1)n+C2+o(1). \begin{gathered} \sum _{1 \leqslant k \leqslant n} k \ln k=\left(n^{2}+n+\frac{1}{6}\right) \frac{\ln n}{2}-\frac{1}{4} n^{2}+C_{1}+o(1), \\ \sum _{1 \leqslant k \leqslant n} \ln k!=\left(n^{2}+2 n+\frac{1}{2}\right) \frac{\ln n}{2}-\frac{3 n^{2}}{4}+\left(\frac{\ln 2 \pi }{2}-1\right) n+C_{2}+o(1) . \end{gathered}
?
Problem 2.18 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 2 "The Degree of Approximation"

Suppose that f(x)f(x) is differentiable and that f′(x)f'(x) is monotone increasing or decreasing to zero as x→∞x \to \infty. Then the following limit exists:

lim⁡n→∞(12f(1)+f(2)+f(3)+⋯+f(n−1)+12f(n)−∫1nf(x) dx)=s. \lim _{n \to \infty } \left(\frac{1}{2} f(1) + f(2) + f(3) + \cdots + f(n-1) + \frac{1}{2} f(n) - \int _{1}^{n} f(x) \, dx\right) = s .

Assume that f′(x)f'(x) increases. Then the two inequalities

18f′(n)<12f(1)+f(2)+f(3)+⋯+f(n−1)+12f(n)−∫1nf(y) dy−s<0 \frac{1}{8} f'(n) < \frac{1}{2} f(1) + f(2) + f(3) + \cdots + f(n-1) + \frac{1}{2} f(n) - \int _{1}^{n} f(y) \, dy - s < 0

can be established. Note the particular cases f(x)=1xf(x) = \dfrac {1}{x}, f(x)=−log⁡xf(x) = -\log x.

?
Problem 2.19 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 2 "The Degree of Approximation"

Assume that f(x)f(x) is differentiable for x≥1x \geq 1 and that f′(x)f'(x) is monotone increasing to ∞\infty as x→∞x \to \infty. Then

12f(1)+f(2)+f(3)+⋯+f(n−1)+12f(n)=∫1nf(x) dx+O[f′(n)]. \frac{1}{2} f(1) + f(2) + f(3) + \cdots + f(n-1) + \frac{1}{2} f(n) = \int _{1}^{n} f(x) \, dx + O\left[f'(n)\right] .

More precisely,

0<12f(1)+f(2)+f(3)+⋯+f(n−1)+12f(n)−∫1nf(x) dx<18f′(n)−18f′(1). 0 < \frac{1}{2} f(1) + f(2) + f(3) + \cdots + f(n-1) + \frac{1}{2} f(n) - \int _{1}^{n} f(x) \, dx < \frac{1}{8} f'(n) - \frac{1}{8} f'(1) .
?

References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.