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Series with non negative real terms

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 $\forall n \in \mathbb{N}, u_n \geq 0$ .

Suppose that there exists a convergent series $\underset{n=0}{\overset{+\infty }{\sum}} v_n$ such that, for a given natural number , $n \geq N \Longrightarrow 0 \leq u_n \leq v_n$ . Then the series $\underset{n=0}{\overset{+\infty }{\sum}} u_n$ is convergent.
Suppose that there exists a divergent series $\underset{n=0}{\overset{+\infty }{\sum}} w_n$ such that, for a given natural number , $n \geq N \Longrightarrow 0 \leq w_n \leq u_n$ . 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 $v_n \geq 0$ .

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.
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.
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

, 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 series with , 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:

If , the series is convergent.
If or $L=+\infty$ , the series is divergent.
If , the test gives no conclusion.

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:

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

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