0
$\begingroup$

$$ a_{n} = 1 + \frac{1}{n^3} $$ Show that the sequence is converges $$ \lim_{n \rightarrow \infty} \left(1 + \frac{1}{1^3}\right)\left(1 + \frac{1}{2^3}\right)\left(1 + \frac{1}{3^3}\right) \ldots \left(1 + \frac{1}{n^3}\right) $$ I know that I should use natural logarithm but I have no clue how. Could you give me a hint?

$\endgroup$
  • $\begingroup$ Is this the sequence {$(1+\frac{1}{n^3})^n$}? $\endgroup$ – Dunka Nov 21 '14 at 5:11
  • $\begingroup$ Just for your curiosity, $a_{\infty}=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{\pi }$. Nice, isn't it ? $\endgroup$ – Claude Leibovici Nov 21 '14 at 8:32
  • $\begingroup$ See convergence criteria for infinite products. $\endgroup$ – Lucian Nov 21 '14 at 9:09
1
$\begingroup$

Hint:

Note that $$\ln \left[ \left(1 + \frac{1}{1^3}\right)\left(1 + \frac{1}{2^3}\right)\left(1 + \frac{1}{3^3}\right) \ldots \left(1 + \frac{1}{n^3}\right)\right] = \ln \left(1 + \frac{1}{1^3}\right) + \cdots + \ln \left(1 + \frac{1}{n^3}\right) $$

Now use the inequality $\ln (1 + x) \leq x$ for all $x > 0$

$\endgroup$
  • $\begingroup$ Why prove that $$ ln( \left(1 + \frac{1}{1^3}\right)\left(1 + \frac{1}{2^3}\right)\left(1 + \frac{1}{3^3}\right) \ldots \left(1 + \frac{1}{n^3}\right)) $$ is converges implicate that the basic sequence is converges? $\endgroup$ – marrandior Nov 21 '14 at 5:21
  • $\begingroup$ Because if we can show $\ln(a_n)$ converges, then $a_n$ converges. My hint can be used to show $\ln(a_n)$ is bounded above. It is also clearly increasing; hence it must have a limit. Therefore $a_n$ has a limit. $\endgroup$ – Simon S Nov 21 '14 at 5:23

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.