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$$\lim_{n \to \infty } \frac{n}{2^{n}}$$

$$ \text{(using L'Hopitals rule)}$$

$$\lim_{n \to \infty } \frac{1}{n2^{n-1}}$$

$$\frac{1}{\infty }=0$$

$\therefore $ Convergent?

Is this the correct solution?

mathreadler
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Idkwoman
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    Of course not. $\sum_{n\geq 1}\frac{1}{n}$ is divergent, even if $\lim_{n\to +\infty}\frac{1}{n}=0$. – Jack D'Aurizio Apr 19 '17 at 10:28
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    No, this is not correct. Observe that $\lim a_n$ is something different than $\lim\sum a_n$. And $[2^x]'=\ln 2\cdot 2^x\neq x\cdot 2^{x-1}$ – Masacroso Apr 19 '17 at 10:29
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    Using L'Hospital's rule for a sequence rather than a function is an abomination! As others have pointed out, this is a case for the ratio test. – Angina Seng Apr 19 '17 at 10:40
  • @Lord you can do that some times, by example if the extension of some $f(n)$ to $\Bbb R$ is strictly monotone then it holds that $$\lim_{n\to\infty}f(n)=\lim_{x\to\infty}f(x)$$ and you can use L'Hôpital rule in the RHS to find the limit. For example in this case the use of L'Hôpital rule is possible, but ofc not directly over $f(n)$ if not over it monotone extension to $\Bbb R$. – Masacroso Apr 19 '17 at 11:05
  • Note that the title asks something entirely different than the problem addressed in the body of the Question. – hardmath Apr 19 '17 at 16:45
  • NO? The title asks whether it is convergent or divergent, and the body just states my approach which could be right or wrong. So the question in the title stands! – Idkwoman Apr 19 '17 at 17:00

3 Answers3

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Hint try comparing to a series if noting $\frac{n}{2^n}<1$

or you can compare to the integral $$\int_{}^\infty \frac{x}{2^x}dx$$

which you can derive with for example integration by parts.

mathreadler
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No. It is true that $\lim_{n\to\infty}\frac{n}{2^n}=0$, but it doesn't follow that the series is convergent.

To show that it is convergent, try using the ratio test.

Especially Lime
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  • Sorry, can you elaborate on the ratio test? Thanks so much! – Idkwoman Apr 19 '17 at 10:32
  • @Idkwoman do you know what ratio test is? Did you apply it? – kingW3 Apr 19 '17 at 10:35
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    Ratio test: if $\big|\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\big|<1$, the series converges. If it is $>1$, the series does not converge. If the limit doesn't exist, or is $1$ or $-1$, the test is inconclusive. – Especially Lime Apr 19 '17 at 10:41
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This is incorrect. In general, if $\sum a_n$ converges, then $a_n \to 0$. However, the converse does not hold. Thus you should find another way. Observe that $$ n < \left(\frac{3}{2}\right)^n $$ for all natural $n$. Then $0< \frac{n}{2^n} < \left(\frac{3}{4}\right)^n$ for all $n$. Since $\sum \left(\frac{3}{4}\right)^n$ converges, $\sum \frac{n}{2^n}$ converges by the comparison test.

In fact, you can find $$ \sum_{n=1}^\infty \frac{n}{2^n}=2. $$

choco_addicted
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