5
$\begingroup$

$$\mathop {\lim }\limits_{n \to \infty } (\sqrt 2 - \root 3 \of 2 )(\sqrt 2 - \root 4 \of 2 )...(\sqrt 2 - \root n \of 2 )$$

I was able easily to find the limit of this series using the squeezing principle, but how can you find it without it? I'm guessing it's an algebra trick :)

$\endgroup$
3
$\begingroup$

I see no trick nor algebra, just the simple observation that for every $k \geq 3$ we have $2^{1/2} > 2^{1/k} > 1$, so that $$ 0 \leq \prod_{k=3}^n \left(2^{1/2}-2^{1/k}\right) \leq \left(\sqrt{2}-1\right)^{n-2} $$ The r.h.s. being a geometric sequence with $0 < \sqrt{2} - 1 < 1$, the sequence converges (very quickly) towards $0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Our sequence is $a_n = \prod_{k=1}^n (2-2^{1/k}).$ Take the log to get $$\log(a_n) = \sum_{k=1}^n \log(2-2^{1/k}).$$ This is a typical thing to do when you have a product. Note that for $k$ large $2-2^{1/k}$ is near $\sqrt{2}-1$ and $\log(\sqrt{2}-1)<0$. Thus the series diverges and $\log(a_n) \to -\infty$. This means $a_n \to 0$.

| cite | improve this answer | |
$\endgroup$

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.