Welcome to our community

Be a part of something great, join today!

Series Integral Comparision

mathworker

Active member
May 31, 2013
118
Is it true that,
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}>\int_{1}^{\infty}\)\(\displaystyle \frac {1}{x}dx\)
explanation is appreciated :)
edit:sorry,lower limit in right hand side is changed from "0" to "1"
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: series integral comparision

Is it true that,
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}>\int_{0}^{\infty}\)\(\displaystyle \frac {1}{x}dx\)
explanation is appreciated :)
No it's not. Your sum is the right-hand estimate for the definite integral, but since the function is decreasing, your sum is an UNDER-estimate.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: series integral comparision

Is it true that,
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}>\int_{0}^{\infty}\)\(\displaystyle \frac {1}{x}dx\)
explanation is appreciated :)
The expressions $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ and$\displaystyle \int_{0}^{\infty} \frac{1}{x}\ dx$ are meaningless because both the series and the integral diverge... may be is true that...

$\displaystyle \sum_{k=1}^{n} \frac{1}{k} > \int_{1}^{n+1} \frac{dx}{x}\ (1)$

Kind regards

$\chi$ $\sigma$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: series integral comparision

Is it true that,
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}>\int_{1}^{\infty}\)\(\displaystyle \frac {1}{x}dx\)
explanation is appreciated :)
edit:sorry,lower limit in right hand side is changed from "0" to "1"
I suspect that the OP is trying to prove that \(\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{n}} \end{align*}\) is divergent. To do this, a simple comparison can be used.

\(\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{n}} &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \dots \\ &> \frac{1}{2} + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \dots \\ &= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots \\ \to \infty \end{align*}\)

Since the harmonic series is greater than this divergent series, the harmonic series is divergent.
 

mathworker

Active member
May 31, 2013
118
Re: series integral comparision

Yeah its meaning less to to compare two infinities, see 10.4 in this author trying two make some sense by comparing area under graphs,is the author wrong doing so?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
No, the author is not wrong to do so. You have interpreted the original integral incorrectly. In this case the author is using the sum as a LEFT endpoint estimate on the integral, as on a decreasing function, you have an OVER estimation. The integral is actually being evaluated between 1 and infinity, not 0 and infinity. So the sum IS greater than the integral in that region, and so can be used to show the divergence of this series.
 

mathworker

Active member
May 31, 2013
118
Actually my question is why did the author use bars for in graph
 

chisigma

Well-known member
Feb 13, 2012
1,704
Actually my question is why did the author use bars for in graph
The task of the author probably is to demonstrate by geometrical evidence that...


$\displaystyle \sum_{k=1}^{n} \frac{1}{k} > \int_{1}^{n+1} \frac{d x}{x} = \ln (n+1)\ (1)$

But if (1) is true, what can we say about the asyntotical behavior of $\displaystyle \sum_{k=1}^{n} \frac{1}{k}$ and $\displaystyle \ln n$?... in the XVIII century the Swiss mathematician Leonhard Euler demonstrated that...


$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n = \gamma\ (2)$

... where $\displaystyle \gamma= .5772...$ is the so called 'Euler's constant'...


Kind regards


$\chi$ $\sigma$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Actually my question is why did the author use bars for in graph
Because each bar has an area that is numerically equal to each term in the sum...
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Re: series integral comparision

I suspect that the OP is trying to prove that \(\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{n}} \end{align*}\) is divergent. To do this, a simple comparison can be used.

\(\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{n}} &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \dots \\ &> \frac{1}{2} + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \dots \\ &= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots \\ \to \infty \end{align*}\)

Since the harmonic series is greater than this divergent series, the harmonic series is divergent.
Very elegant! I'll have to remember this one.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: series integral comparision

Very elegant! I'll have to remember this one.
This 'very elegant' prove that the harmonic series diverges was found in the 14th century by the French mathematician, economist, phisician, astronomer, astrologist, philosoph and theologian Nicolas d'Oresme, bishop of Lisieaux, one of the most original a versatile minds of the Middle Age...

Kind regards

$\chi$ $\sigma$