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Does the following series converge ?

$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}+\ldots$$

Let, $u_{n}=\frac{1}{\sqrt{n}}$

$\lim\limits_{n\to\infty}{u_{n}}=\lim\limits_{n\to\infty}\frac{1}{\sqrt{n}}$

$\lim\limits_{n\to\infty}{u_{n}}= 0$

Julien
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HOLYBIBLETHE
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  • See theorem 3.20 in Walter Rudin's Principles of Mathematical Analysis . –  Apr 11 '13 at 06:03

6 Answers6

6

$$\sum_{n = 1}^\infty \frac{1}{\sqrt{n}} = \sum_{n = 1}^\infty \frac 1{n^{(1/2)}}$$

By the $p-\text{series test},\;$ if the exponent $p$ of $n$ in the denominator is $p \leq 1$, the series diverges. Here, that exponent $p = 1/2 \lt 1$, hence, the series diverges.

amWhy
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5

$1+\dfrac{1}{2}+\dfrac{1}{3}+\cdot\cdot\cdot+\dfrac{1}{n}$ is diverges when $n\to\infty$, $\dfrac{1}{\sqrt{n}}>\dfrac{1}{n}$, so the series is diverges when $n\to\infty$.

Martin Sleziak
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chenbai
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2

Hint: Use the integral test,

$$ \int_{1}^{\infty}\frac{1}{\sqrt{x}} dx. $$

Added: Here is the main result.

Theorem: Suppose $f$ is continuous, positive, decreasing function on $[1,\infty)$ and let $a_n=f(n)$. Then

$(a)$ if $\int_{1}^{\infty}f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n $ is convergent.

$(b)$ if $\int_{1}^{\infty}f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n $ is divergent.

Note: Now, you can see, for any series of the form $\sum_{n=1}^{\infty}\frac{1}{n^p}$, we can find the condition on $p$ for which the series converges or diverges by considering the integral

$$ \int_{1}^{\infty}\frac{1}{x^p} dx. $$

Mhenni Benghorbal
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1

This series diverges from the $p$-test. If you're unfamiliar with this approach to the problem, check this link: http://www.sosmath.com/calculus/improper/testconv/testconv.html. This is a more elementary method. Rather than using the integral test like Sean Gomes suggested.

MITjanitor
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    Isn't the "p-test" typically proven using the integral test? – goonfiend Apr 11 '13 at 05:20
  • @SeanGomes Good one. But the "p-test" if you know it, is the easiest way to evaluate this integral. If the OP would have said his prior knowledge, my answer would have been different. – MITjanitor Apr 11 '13 at 08:45
0

No.

In general, the $n$-th term of a sequence tending to zero just not imply that the corresponding infinite series converges.

One way of proving divergence in this example is by using the integral test.

goonfiend
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0

General purpose tests are really useful for such problems (and there are answers with those).

Here is a different way of proving divergence.

$$\sqrt{k+1} - \sqrt{k} = \frac{1}{\sqrt{k+1} + \sqrt{k}} \le \frac{1}{2\sqrt{k}}$$

Thus

$$\sqrt{n+1} - 1 = \sum_{k=1}^{n} (\sqrt{k+1} - \sqrt{k}) \le \sum_{k=1}^{n} \frac{1}{2\sqrt{n}}$$

and so

$$\sum_{k=1}^{n} \frac{1}{\sqrt{n}} \ge 2(\sqrt{n+1} - 1)$$

Aryabhata
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