4
$\begingroup$

Does the following series converge ?

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

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

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

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

Improve this question
$\endgroup$
  • $\begingroup$ See theorem 3.20 in Walter Rudin's Principles of Mathematical Analysis . $\endgroup$ – user70532 Apr 11 '13 at 6:03
6
$\begingroup$

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

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.

Improve this answer
$\endgroup$
  • $\begingroup$ The $p$-series is a consequence of the integral test. $\endgroup$ – Mhenni Benghorbal Apr 11 '13 at 5:59
5
$\begingroup$

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

Improve this answer
$\endgroup$
2
$\begingroup$

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

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

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Isn't the "p-test" typically proven using the integral test? $\endgroup$ – Goonfiend Apr 11 '13 at 5:20
  • $\begingroup$ @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. $\endgroup$ – MITjanitor Apr 11 '13 at 8:45
0
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

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

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

Improve this answer
$\endgroup$
  • $\begingroup$ @ Aryabhata : You are right. Nature always provides atleast two ways to solve a problem at hand. $\endgroup$ – HOLYBIBLETHE Apr 12 '13 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.