The Test of Positive Series

When handling problems of infinite series, we always ask ourselves two important questions:

• Does it converge?
• If converges, what is its sum?

Some tests can help us to find the answer.

The Integral Test

Let ${\displaystyle f}$ be a continuous, positive, nonincreasing function, and suppose that ${\displaystyle a_n=f(n)}$ for all positive integers ${\displaystyle n}$, then the infinite series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$

converges if and only if the improper integral

${\displaystyle {\int }_1^{\mathrm{\infty }}f(x)dx}$

converges.

If ${\displaystyle f}$ is decreasing, then ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ and ${\displaystyle {\int }_1^{\mathrm{\infty }}f(x)dx}$ both converge or both diverge.

Ordinary Comparison Test

For ${\displaystyle 0<a_n<b_n}$, if:

• ${\displaystyle \sum b_n}$ converges, so does ${\displaystyle a_n}$.
• ${\displaystyle \sum a_n}$ diverges, so does ${\displaystyle b_n}$.

This is a very helpful way to determine if a series converges or diverges. We usually use geometric or p series as the upper or lower bound to prove the given series's convergence or divergence.

e.g.

For ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{n}{5n^2-4}}$, does it converge or diverge?

sol.

Since ${\displaystyle \frac{n}{5n^2-4}<\frac{n}{5n^2}=\frac{1}{5n}}$

And we know that ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{5n}}$ diverges since it is a harmonic series. So the given series must diverge because its lower bound diverges.

Limit Comparison Test

For ${\displaystyle a_n\ge 0,b_n>0}$, and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=L}$, if:

• ${\displaystyle 0<L<\mathrm{\infty }}$, then ${\displaystyle a_n}$ and ${\displaystyle b_n}$ converge or diverge together.
• ${\displaystyle L=0}$, then ${\displaystyle a_n}$ converges if and only if ${\displaystyle b_n}$ converges.

This is a bit like the determination of equivalent infinitesimals. It helps especially when ${\displaystyle \frac{a_n}{b_n}}$ is the series in the form:

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{b_0{n}^q+b_1{n}^{q-1}+\cdots +bq}{a_0{n}^p+a_1{n}^{p-1}+\cdots +ap}}$

Which is actually a rational but discontinuous function.

When:

• ${\displaystyle p-q>1}$, the series of the ratio converges.
• ${\displaystyle p-q\le 1}$, the series of the ratio diverges.

It is very similar to the way of finding equivalent infinitesimals. When the denominator's exponent is larger then that of the numerator, the ratio approaches to ${\displaystyle 0}$ when ${\displaystyle n}$ grows bigger and bigger.

Ratio Test

For ${\displaystyle a_n>0}$, and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}=\rho }$, if:

• ${\displaystyle \rho <1}$, then ${\displaystyle \sum a_n}$ converges.
• ${\displaystyle \rho >1}$, then ${\displaystyle \sum a_n}$ diverges.
• ${\displaystyle \rho =1}$, the test does not show anything, we can choose another test.

The way this test works is obvious, which shows the trend of change of the sequence. If the next item is always smaller than the previous one, since the series is always positive for all ${\displaystyle n>N}$, it will moving closing and closing to ${\displaystyle 0}$, which means that the series converges.

Root Test

For ${\displaystyle a_n{\ge }_0}$, and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{a_n}=\rho }$, if:

• ${\displaystyle \rho >1}$, then ${\displaystyle \sum a_n}$ converges.
• ${\displaystyle \rho >1or\rho =\mathrm{\infty }}$, then ${\displaystyle \sum a_n}$ diverges.
• ${\displaystyle \rho =1}$, the test does not show anything, we can choose another test.

This test is useful when we are facing series of exponent functions. It is similar to the ratio test.