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Upper bound on alternating harmonic series

Harmonic Numbers series IDouble series of Harmonic NumbersProof of an upper bound for $sum_{i=c}^{infty} frac{1}{i^k}$Harmonic Sum and cardinality of continuumShow monotonicity of an Alternating SeriesUpper bound on a series: $sum_{k = M}^{infty} (a/sqrt(k))^k$Upper bound on a series of integralsAlternating harmonic series containing floor functionrearrangement of the alternating harmonic seriesUpper bound for the probability of violating a set of conditions

0

2

$begingroup$

Let $a > 0$. Is the following statement true:
begin{align}
ln(2) = sum_{n=1}^infty{frac{(-1)^{n+1}}{n}} > sum_{n=1}^infty{frac{(-1)^{n+1}}{n+a}}
end{align}
From my intuition the statement is true, because in absolute terms the numbers on the right hand side are smaller

sequences-and-series upper-lower-bounds

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asked Mar 13 at 16:19

cluelessclueless

307111

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0

2

$begingroup$

Let $a > 0$. Is the following statement true:
begin{align}
ln(2) = sum_{n=1}^infty{frac{(-1)^{n+1}}{n}} > sum_{n=1}^infty{frac{(-1)^{n+1}}{n+a}}
end{align}
From my intuition the statement is true, because in absolute terms the numbers on the right hand side are smaller

sequences-and-series upper-lower-bounds

share|cite|improve this question

asked Mar 13 at 16:19

cluelessclueless

307111

$endgroup$

add a comment | 

$begingroup$

Let $a > 0$. Is the following statement true:
begin{align}
ln(2) = sum_{n=1}^infty{frac{(-1)^{n+1}}{n}} > sum_{n=1}^infty{frac{(-1)^{n+1}}{n+a}}
end{align}
From my intuition the statement is true, because in absolute terms the numbers on the right hand side are smaller

sequences-and-series upper-lower-bounds

share|cite|improve this question

asked Mar 13 at 16:19

cluelessclueless

307111

$endgroup$

share|cite|improve this question

asked Mar 13 at 16:19

cluelessclueless

307111

share|cite|improve this question

asked Mar 13 at 16:19

cluelessclueless

307111

asked Mar 13 at 16:19

cluelessclueless

307111

asked Mar 13 at 16:19

cluelessclueless

307111

2 Answers
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1

$begingroup$

The argument based on absolute values may not be correct since the terms may alternatively go below or above each other. A safer way naturally is to remove the oscillation in the terms:
$$x_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n}\=sum_{k=0}^infty {1over (2k+1)(2k+2)}}$$
$$y_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n+a}\=sum_{k=0}^infty {1over (2k+1+a)(2k+2+a)}}$$since for $a>0$ we can say $${1over (2k+1+a)(2k+2+a)}<{1over (2k+1)(2k+2)}$$by term by term summation we can conclude what we want.

share|cite|improve this answer

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

$endgroup$

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0

$begingroup$

We have

$$sum_{n=1}^infty left( frac{(-1)^{n+1}}{n} - frac{(-1)^{n+1}}{n+a} right) = sum_{n=1}^infty (-1)^{n+1} left( frac{1}{n} - frac{1}{n+a} right) = a sum_{n=1}^infty frac{(-1)^{n+1}}{n(n+a)}$$

This is greater than zero because this is an alternating series and $frac{1}{n(n+a)}$ is strictly decreasing.

share|cite|improve this answer

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

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add a comment | 

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1

$begingroup$

The argument based on absolute values may not be correct since the terms may alternatively go below or above each other. A safer way naturally is to remove the oscillation in the terms:
$$x_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n}\=sum_{k=0}^infty {1over (2k+1)(2k+2)}}$$
$$y_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n+a}\=sum_{k=0}^infty {1over (2k+1+a)(2k+2+a)}}$$since for $a>0$ we can say $${1over (2k+1+a)(2k+2+a)}<{1over (2k+1)(2k+2)}$$by term by term summation we can conclude what we want.

share|cite|improve this answer

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

$endgroup$

add a comment | 

1

$begingroup$

The argument based on absolute values may not be correct since the terms may alternatively go below or above each other. A safer way naturally is to remove the oscillation in the terms:
$$x_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n}\=sum_{k=0}^infty {1over (2k+1)(2k+2)}}$$
$$y_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n+a}\=sum_{k=0}^infty {1over (2k+1+a)(2k+2+a)}}$$since for $a>0$ we can say $${1over (2k+1+a)(2k+2+a)}<{1over (2k+1)(2k+2)}$$by term by term summation we can conclude what we want.

share|cite|improve this answer

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

$endgroup$

add a comment | 

$begingroup$

The argument based on absolute values may not be correct since the terms may alternatively go below or above each other. A safer way naturally is to remove the oscillation in the terms:
$$x_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n}\=sum_{k=0}^infty {1over (2k+1)(2k+2)}}$$
$$y_n{=sum_{n=1}^{infty}{(-1)^{n+1}over n+a}\=sum_{k=0}^infty {1over (2k+1+a)(2k+2+a)}}$$since for $a>0$ we can say $${1over (2k+1+a)(2k+2+a)}<{1over (2k+1)(2k+2)}$$by term by term summation we can conclude what we want.

share|cite|improve this answer

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

$endgroup$

share|cite|improve this answer

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

answered Mar 13 at 17:41

Mostafa AyazMostafa Ayaz

17.2k31039

0

$begingroup$

We have

$$sum_{n=1}^infty left( frac{(-1)^{n+1}}{n} - frac{(-1)^{n+1}}{n+a} right) = sum_{n=1}^infty (-1)^{n+1} left( frac{1}{n} - frac{1}{n+a} right) = a sum_{n=1}^infty frac{(-1)^{n+1}}{n(n+a)}$$

This is greater than zero because this is an alternating series and $frac{1}{n(n+a)}$ is strictly decreasing.

share|cite|improve this answer

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

$endgroup$

add a comment | 

0

$begingroup$

We have

$$sum_{n=1}^infty left( frac{(-1)^{n+1}}{n} - frac{(-1)^{n+1}}{n+a} right) = sum_{n=1}^infty (-1)^{n+1} left( frac{1}{n} - frac{1}{n+a} right) = a sum_{n=1}^infty frac{(-1)^{n+1}}{n(n+a)}$$

This is greater than zero because this is an alternating series and $frac{1}{n(n+a)}$ is strictly decreasing.

share|cite|improve this answer

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

$endgroup$

add a comment | 

$begingroup$

We have

$$sum_{n=1}^infty left( frac{(-1)^{n+1}}{n} - frac{(-1)^{n+1}}{n+a} right) = sum_{n=1}^infty (-1)^{n+1} left( frac{1}{n} - frac{1}{n+a} right) = a sum_{n=1}^infty frac{(-1)^{n+1}}{n(n+a)}$$

This is greater than zero because this is an alternating series and $frac{1}{n(n+a)}$ is strictly decreasing.

share|cite|improve this answer

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

$endgroup$

share|cite|improve this answer

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423

answered Mar 13 at 16:29

Haris GusicHaris Gusic

2,738423