how to show that series diverges

[evinda]

Hi!!
How can I show that the series \(\displaystyle \sum \frac{1}{2n+1} \) diverges???

 

[lookagain]

Hi!!
How can I show that the series \(\displaystyle \sum \frac{1}{2n+1} \) diverges???

You mean this, correct?


\(\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{1}{2n+1} \ \ \ \ or \ \ possibly \ \ this \ \ \ \ \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n+1}\)



If so, do you know that the following diverges?


\(\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}\)

 

[evinda]

You mean this, correct?


\(\displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{1}{2n+1} \ \ \ \ or \ \ possibly \ \ this \ \ \ \ \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n+1}\)



If so, do you know that the following diverges?


\(\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}\)

Yes,I mean the second one \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n+1}\).

\(\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\) ,that diverges because it is the harmonic series.Right?

 

[lookagain]

Yes,I mean the second one \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n+1}\).

\(\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\) ,that diverges because it is the harmonic series.Right?

Yes, to the right of the "1/2" is the harmonic series.

This is information so far toward an answer.

 

[pka]

\(\displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}=\frac{1}{2}\displaystyle\sum_{n=1}^{
\infty} \frac{1}{n}\) ,that diverges because it is the harmonic series.Right?

Actually is did no matter where you begin your index.
\(\displaystyle (\forall k\in \mathbb{Z})\left[\displaystyle \sum_{n=k}^{\infty} \frac{1}{2n+1}\right]\) diverges.

And yes it is like the harmonic series.

 

[Romsek]

Actually is did no matter where you begin your index.
\(\displaystyle (\forall k\in \mathbb{Z})\left[\displaystyle \sum_{n=k}^{\infty} \frac{1}{2n+1}\right]\) diverges.

And yes it is like the harmonic series.

If we want to be really pedantic about it we can note that

\(\displaystyle
\frac{1}{2 k+1}>\frac{1}{2 k+2}\) for k>0

\(\displaystyle \frac{1}{2 k+1}>\frac{1}{2 (k+1)}\) for k>0

\(\displaystyle
\sum _{k=0}^{\infty } \frac{1}{2 k+1}>\frac{1}{2} \sum _{k=0}^{\infty }
\frac{1}{k+1}=\frac{1}{2} \sum _{k=1}^{\infty } \frac{1}{k}
\)

We recognize the last term as the half the harmonic series, which diverges, and since our series is larger term by term than a divergent series it must diverge as well.

 

[pka]

If we want to be really pedantic about it we can note that
\(\displaystyle \frac{1}{2 k+1}>\frac{1}{2 k+2}\) for k>0

\(\displaystyle \frac{1}{2 k+1}>\frac{1}{2 (k+1)}\) for k>0

We recognize the last term as the half the harmonic series, which diverges, and since our series is larger term by term than a divergent series it must diverge as well.

If you really want to be pedantic, you should know that has little to do with convergence or divergence.

The saying goes: "only the tail matters". If from some index on the series convergence or divergence then the same is true of the whole series. A series does what almost all (all but a finite collection) of its terms do.

 

[daon2]

evinda, here is an approach that you might think about, but not expect to look at in your classroom/studies:

\(\displaystyle \displaystyle\sum_{n=1}^{\infty}\dfrac{1}{ \ 2n + 1 \ } \ = \)



\(\displaystyle \displaystyle\sum_{n=1}^{\infty}\dfrac{1}{ \ 2n \ } \ - \ \displaystyle\sum_{n=1}^{\infty}\dfrac{1}{ \ 4n^2 + 2n \ } \ =\)


Then, for an emphasis on the harmonic series and the p-series, we can continue it as:



\(\displaystyle \dfrac{1}{2}\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{ \ n \ } \ - \ \frac{1}{4}\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{ \ n^2 + \frac{1}{2}n \ } \)

This is not allowed.

\(\displaystyle \displaystyle \sum (a_n+b_n) = \sum a_n+ \sum b_n\) only if both sums on the right converge.

My example here had an error, but a student would be docked for asserting that equality

 

[lookagain]

I would rather suggest as an alternative to just using the integral test of convergence to show that it is divergent.