Convergence Criteria

Remark 6.2.1  
The most commonly used criteria to determine the convergence of series with positive terms are:

• Comparison Test for Convergence: Suppose that $\sum a_k$ and $\sum b_k$ are series with positive terms. If $\sum b_k$ is convergent and $a_k \leq b_k$ for any $k$, then $\sum a_k$ is also convergent.

  Example: $\frac{1}{\ln(k) 2^k} \leq \frac{1}{2^k}$ and $\sum_{k=0}^{\infty}\frac{1}{2^k} = 2$, so $\sum_{k=0}^{\infty}\frac{1}{\ln(k) 2^k}$ is also convergent. Note that this property only gives us the fact if a series is convergent or not and not what the actual sum is.

• Comparison Test for Divergence: Suppose that $\sum a_k$ and $\sum b_k$ are series with positive terms. If $\sum b_k$ is divergent and $a_k \geq b_k$ for any $k$, then $\sum a_k$ is also divergent.

  Example: $2^k \leq 2^k+\frac{1}{k!}$ and $\sum 2^k$ is divergent, so $\sum 2^k + \frac{1}{k!}$ is also divergent.

• Limit Comparison Test: Suppose that $\sum a_k$ and $\sum b_k$ are series with positive terms. If

  $$\lim_{k \to \infty}\frac{a_k}{b_k} = c$$ (6.2.1)

  
  
  where $c$ is a finite number and $c>0$, then either both series converge or both series diverge.

  Examples: Look at $a_k = \frac{1}{2^k}$ and $b_k = \frac{1}{2^k-1}$, the limit comparison gives

  $$\lim_{k \to \infty}\frac{a_k}{b_k} = \lim_{k \to \infty}\frac{2^k}{2^k-1} = \lim_{k \to \infty}\frac{1}{1-\left(\frac{1}{2}\right)^k} = 1 > 0$$ (6.2.2)

  
  

  Because $\sum \frac{1}{2^k}$ is a convergent series, then so is $\sum \frac{1}{2^k-1}$.
  Note that the condition $c>0$ is a strict inequality. This is necessary. Take e.g. $a_k = \frac{1}{k^2}$ (called the hyper harmonic series of second order) and take $b_k = \frac {1}{k}$ (called the harmonic series), then the limit is

  $$\lim_{k \to \infty}\frac{a_k}{b_k} = \lim_{k \to \infty}\frac{\frac{1}{k^2}}{\frac{1}{k}} = \lim_{k \to \infty}\frac{1}{k} = 0$$

  
  

  But, on the other hand, the harmonic series is divergent while the hyper harmonic series is convergent. Neither fact is that easy to see, hence the following two sections to discuss this in detail.

Example 6.2.2 (Harmonic Series)  
Establishing the divergence of the harmonic series boils down to seeing that the sum

$$S_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \ldots$$ (6.2.3)

tails off to infinity. To see this, observe the following for $S_8$:

$$\begin{aligned}S_8 &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{... ...) \\ &> 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \end{aligned}$$

In general, this grouping of terms leads to

$$S_{2^n} > 1 + \frac{n}{2}$$ (6.2.5)

and thus

$$\lim_{n \to \infty}S_{2^n} > 1 + \lim_{n \to \infty}\frac{n}{2} = \infty$$ (6.2.6)

Example 6.2.3 (Hyperharmonic Series)  
It is of considerable technical difficulty to establish the value of the sum

$$S_n = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \ldots$$ (6.2.7)

it is however fairly easy to establish that the sum does converge:

$$\begin{aligned}S_n &= 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac... ...1}{8} + \frac{1}{16} + \frac{1}{32} +\ldots \\ &= 2 \end{aligned}$$

This little computation only proves that the hyper harmonic series of order $2$ is indeed convergent, but not that it's actual value is $\frac{\pi^2}{6}$ (calculating this value would really be going off on a tangent).

Theorem 6.2.4 (Integral Test)  
Let $f$ be a continuous, positive, decreasing function on $[1,\infty[$ and let $a_k = f(k)$. Then

$$\sum_{k=0}^{\infty}a_k \hbox{ is convergent iff } \int_1^{\infty} f(x) dx < \infty$$ (6.2.9)

Proof.
As $f$ is a decreasing function, for any $x$ in $[k,k+1]$, we know $f(k+1) \leq f(x)$. So we have

$$f(k+1) \I1_{[k,k+1[} (x) \leq \int_k^{k+1} f(x) dx$$ (6.2.10)

Then we have

$$\sum_{k=1}^n f(k+1) \I1_{[k,k+1[} (x) \leq \int_1^n f(x) dx \leq \int_1^{\infty} f(x) dx$$ (6.2.11)

As this is true for any $n$, this means

$$\sum_{k=0}^{\infty}a_{k+1} = \sum_{k=1}^{\infty} f(k+1) \I1_{[k,k+1[} (x) \leq \int_1^{\infty} f(x) dx$$ (6.2.12)

So if the integral is finite, the sum converges, which means we have proven one direction. Formula (6.2.10) might look a bit intimidating, but all we are doing is simply taking the lower Riemann sums:

180pt

On the other hand, we can follow the same reasoning for the upper Riemann sums. Because $f$ is a decreasing function, for any $x$ in $[k,k+1]$, we know $f(x) \leq f(k)$. So we have

$$\sum_{k=0}^nf(k+1) \I1_{[k,k+1[} (x) \leq \int_1^k f(x) dx$$ (6.2.13)

Which then leads to

$$\int_1^n f(x) dx \leq \sum_{k=1}^n f(k) \I1_{[k,k+1[} (x)$$ (6.2.14)

Because this is true for any $n$, this means

$$\int_1^{\infty} f(x)dx \leq \sum_{k=1}^{\infty} f(k) \I1_{[k,k+1[} = \sum_{k=0}^{\infty}a_k$$ (6.2.15)

So if the sum is finite, the integral is finite and therefore convergence of the sum means convergence of the integral. Formula (6.2.13) might look a bit intimidating, but again, it is a very natural idea, namely taking Riemann upper sums:

180pt

Note that in either case we are estimating the integral with only part of the actual sum. This is perfectly all right, because dropping one or two terms of a series does not make any difference to the fact that the series is convergent or divergent. $\qedsymbol$

Example 6.2.5  
This theorem is very handy to prove the convergence of the hyper harmonic series, not just the hyper harmonic series of order $2$, but of any order. Take $p>1$, not necessarily a natural number, even $p=1.0000000000000001$ will do just fine. Note that the function $f(x)=\frac{1}{x^p}$ is continuous, positive and decreasing over $[1,\infty[$. So we can apply the integral test and thus:

$$\sum_{k=0}^{\infty}\frac{1}{k^p} < \infty$$ (6.2.16)

for any $p>1$ because

$$\int_1^{\infty} \frac{1}{x^p} dx = \left[ \frac{x^{1-p}}{1-p} \right]_1^{\infty} = \frac{1}{p-1} < \infty$$ (6.2.17)

On the other hand,

$$\int_1^{+\infty} \frac{1}{x^p} dx = +\infty$$ (6.2.18)

for $p \leq 1$, so

$$\sum_{k=0}^{\infty}\frac{1}{k^p} < \infty$$ (6.2.19)

diverges for $p \leq 1$. Note that this is essentially the hyperharmonic series version of an improper integrals result, namely (5.3.7).

Theorem 6.2.6 (Alternating Series Test)  
If the alternating series

$$\sum_{k=0}^{\infty}(-1)^{k-1} b_k$$ (6.2.20)

with $b_k > 0$ is decreasing (i.e. $b_{k+1} \leq b_k$) and $\lim_{k \to \infty}b_k = 0$, then the series is convergent.

Proof.
Note that for all even numbered sums, we can proceed as follows:

$$\begin{aligned}S_{2n} &= b_1 - b_2 + b_3 - b_4 + b_5 - b_6 + \ldots - b_{2n} \\ &= S_{2n-2} + (b_{2n-1} - b_{2n}) \\ \end{aligned}$$

Because $b_{2n-1} \geq b_{2n}$, we have that $S_{2n} \geq S_{2n-2}$ for any $n$. On the other hand, $S_{2n} \leq b_1$, so we know $(S_{2n})_n$ to be a bounded increasing positive series. Therefore, $S_{2n}$ converges. For $S_{2n+1}$, the situation is no different:

$$\lim_{n \to \infty}S_{2n+1} = \lim_{n \to \infty}S_{2n} + \lim_{n \to \infty}b_{2n+1} = \lim_{n \to \infty}S_{2n}$$ (6.2.22)

So $(S_n)_n$ is convergent, for any $n$, even or odd. $\qedsymbol$

Example 6.2.7  
The alternating harmonic series

$$S_n = \sum_{k=0}^{\infty}\frac{(-1)^k}{k}$$ (6.2.23)

satisfies the condition for the alternating series test and is therefore convergent.

Definition 6.2.8 (Absolute Convergence)  
A series $\sum a_k$ is called absolutely convergent if the series $\sum \vert a_k\vert$ converges.

Example 6.2.9  
Clearly every absolutely convergent series is convergent because

$$\sum_{k=0}^{\infty}a_k \leq \sum_{k=0}^{\infty}\vert a_k \vert$$ (6.2.24)

The converse is definitely no true. Take e.g. the alternating harmonic series. This series is convergent (6.2.7) but taking the absolute value of each term would make it the harmonic series, which is divergent (6.2.2).

Remark 6.2.10  
For absolute convergence, we have the following two main tests:

• The Ratio Test :

  If $$\lim_{k \to \infty}\left\vert \frac{a_{k+1}}{a_k} \right\vert < 1$$, then the series is absolutely convergent.
  

  If $$\lim_{k \to \infty}\left\vert \frac{a_{k+1}}{a_k} \right\vert > 1$$ (including infinity), then the series is divergent.
  

  Note that the ratio test gives absolutely no information if the limit happens to be $1$.

• The Root Test :

  If $$\lim_{k \to \infty}\sqrt[k]{\vert a_k\vert} < 1$$, then the series is absolutely convergent.
  

  If $$\lim_{k \to \infty}\sqrt[k]{\vert a_k\vert} > 1$$ (including infinity), then the series is divergent.
  

  Again, the root test gives absolutely no information if the limit happens to be $1$.