Series with non negative real terms.

We recall here the most important tests for convergence, which will be useful in the next sections. There exist other tests; the interested reader can see them in the Calculus Tutorial.

$\mathbb{C}$

Proposition 7.1.7 (Direct Comparison)   Let $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ be a series of real numbers, such that $ \forall n \in \mathbb{N}, u_n \geq 0$ .
  1. Suppose that there exists a convergent series $ \underset{n=0}{\overset{+\infty }{\sum}} v_n$ such that, for a given natural number $ N$ , $ n \geq N \Longrightarrow 0 \leq u_n \leq v_n$ . Then the series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ is convergent.
  2. Suppose that there exists a divergent series $ \underset{n=0}{\overset{+\infty }{\sum}} w_n$ such that, for a given natural number $ N$ , $ n \geq N \Longrightarrow 0 \leq w_n \leq u_n$ . then the series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ is divergent.

Example 7.1.8   Consider the series $ \underset{n=0}{\overset{+\infty }{\sum}} e^{-n^2}$ . This is a series with positive terms.

For $ n \geq 1$ , we have $ e^{-n^2} \leq e^{-n}$ . By the integral test (consult the Calculus Tutorial), the series $ \underset{n=0}{\overset{+\infty }{\sum}} e^{-n}$ is convergent; thus the given series $ \underset{n=0}{\overset{+\infty }{\sum}} e^{-n^2}$ is convergent.

Proposition 7.1.9 (Limit Comparison)   Let $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ and $ \underset{n=0}{\overset{+\infty }{\sum}} v_n$ be two series of real numbers, such that $ \forall n \in \mathbb{N}, u_n \geq 0$ and $ v_n \geq 0$ .
  1. If there exists a positive real number L such that $ \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_n}{v_n} = L$ , then the two series either are both convergent or are both divergent.
  2. If $ \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_n}{v_n} =0$ and $ \underset{n=0}{\overset{+\infty }{\sum}} v_n$ is convergent, then $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ is convergent.
  3. If $ \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_n}{v_n} = + \infty$ and $ \underset{n=0}{\overset{+\infty }{\sum}} v_n$ is divergent, then $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ is divergent.

Example 7.1.10   Consider the series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ , where $ u_n=\frac {100n+345}{0.n^4+1}$ . For arbitrarily large $ n$ , we have $ u_n \equiv \frac {100n}{0.4n^4}=\frac {100}{0.4n^3} = \frac {250}{n^3}$ .

Take $ v_n= \frac {250}{n^3}$ . Then:

$\displaystyle \frac {u_n}{v_n} = \frac {\frac {100n+345}{0.n^4+1}}{\frac {250}{n^3}} =\frac {.5n^3(20n+69)}{n^4+2.5}$    

Thus:

$\displaystyle \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_n}{v_n} =...
...{n^4+2.5} = \underset{n \rightarrow + \infty }{\text{lim}} \frac{10n^4}{n^4}=10$    

The series $ \underset{n=0}{\overset{+\infty }{\sum}} \frac {250}{n^3}$ is a Riemann $ p-$ series with $ p>1$ , thus it is convergent; therefore, by Thm 1.9, the given series is convergent.

Proposition 7.1.11 (Ratio Test; d'Alembert's Test)   Let $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ such that $ \forall n \in \mathbb{N}, u_n \geq 0$ . Suppose that:

$\displaystyle \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_{n+1}}{u_n} = L.$    

Then:

Example 7.1.12   Take the series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ with general term $ u_n= \frac {3^n -7}{2^n}$ . We have:

$\displaystyle \frac {u_{n+1}}{u_n}= \frac {\frac {3^{n+1} -7}{2^+{n+1}}}{\frac ...
...{3^n -7} \cdot \frac{2^n}{2^+{n+1}} =\frac 12 \cdot \frac {3^{n+1} -7}{3^n -7}.$    

Thus:

$\displaystyle \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_{n+1}}{u_...
... \underset{n \rightarrow + \infty }{\text{lim}} \frac {3^{n+1}}{3^n} =\frac 32.$    

As this limit is more than 1, the given series is divergent.

Example 7.1.13   The series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ with general term $ u_n= \frac{5^n (n!)^2}{(2n)!}$ is given. We have:

$\displaystyle \frac {u_{n+1}}{u_n}= \frac { \frac{5^{n+1} [(n+1)!]^2}{[2(n+1)])!}}{ \frac{5^n (n!)^2}{(2n)!}} =\frac {5 (n+1)}{2(2n+1)}$    

Thus:

$\displaystyle \underset{n \rightarrow + \infty }{\text{lim}} \frac {u_{n+1}}{u_...
... \underset{n \rightarrow + \infty }{\text{lim}} \frac {n+1}{2(2n+1)} = \frac 54$    

By Thm 1.11, the given series diverges.

Noah Dana-Picard 2007-12-24