MHB How can we prove that if a sequence is decreasing and has a sum of infinity, then the limit of n times the nth term of the sequence is zero?"

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Here is the first problem for 2015! ______________

Show that if $(a_n)$ is a decreasing sequence of positive real numbers such that $\sum_{n = 1}^\infty a_n$ converges, then $\lim_{n\to \infty} na_n = 0$. As a consequence, prove that the series $\sum_{n = 1}^\infty (n^{1/n} - 1)$ diverges.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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Fallen Angel gets honorable mention for correctly answering the second part of the problem. You can find my solution below.

Spoiler

Let $s_n$ be the sequence of partial sums of $\sum_{n = 1}^\infty a_n$. Let $t_n := na_n$. Since $a_n$ is decreasing,

$$s_{2n} - s_n = a_{n+1} + \cdots + a_{2n} \ge na_{2n} = \frac{1}{2}t_{2n}$$

and similarly

$$s_{2n+1} - s_n \ge (n+1)a_{2n+1} = \frac{1}{2}t_{2n+1} - a_{2n+1}.$$

Since $\sum a_n$ converges, $\lim a_n = 0$. Therefore, since $s_n$ converges, the above inequalities imply $\lim_{n\to \infty} t_{2n} = 0 = \lim_{n\to \infty} t_{2n+1}$. Therefore, $\lim_{n\to \infty} t_n = 0$.

In the case of the example, the sequence $a_n := n^{1/n} - 1$ is positive decreasing sequence such that $\lim na_n \neq 0$. Hence, by the theorem, $\sum_{n = 1}^\infty (n^{1/n} - 1)$ diverges.

 

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Hi MHB,

I sincerely apologize for overlooking some of the POTW solutions for this week's POTW. Thanks to anemone for notifying me! After reviewing the solutions, I found that Opalg has correctly solved this week's problem. You can read his solution below.

Spoiler

Given $\varepsilon>0$, choose $M$ so that $\sum_{n=M}^\infty a_n <\varepsilon/2$. Then choose $N>M$ so that $n>N\;\Rightarrow\; a_n < \varepsilon/(2M)$. If $n>N$ then $$\begin{aligned} na_n &= Ma_n + (n-M)a_n \\ &< Ma_n + \sum_{r=M+1}^n a_r \\ &< \varepsilon/2 + \varepsilon/2 = \varepsilon . \end{aligned}$$ Since $\varepsilon$ is arbitrary, it follows that $\lim_{n\to\infty}na_n = 0.$

Now let $a_n = n^{1/n} - 1$. This gives a decreasing sequence of real numbers. So if $\sum a_n$ converges then $na_n \to0$ as $n\to\infty.$ But $na_n = f(1/n)$, where $f(x) = \dfrac{\frac1{x^x} - 1}x$, so it would help to know whether $f(x)\to0$ as $x\searrow0$.The function $x^x$ converges to $1$ as $x\searrow0$. Its derivative $(1+\ln x)x^x$ goes to $-\infty$ as $x\searrow0$, so its graph has a vertical tangent at the point $(0,1)$. The reciprocal of this function, $g(x) = 1/x^x$ (with $g(0)=1$), also has a vertical tangent at $(0,1)$. Therefore the Newton quotient $\dfrac{g(x) - g(0)}x$ must go to $+\infty$ as $x\searrow0$. But $\dfrac{g(x) - g(0)}x$ is exactly the expression $f(x) = \dfrac{\frac1{x^x} - 1}x$ from the previous paragraph. Therefore $f(x) \to+\infty$ as $x\searrow0$. So it certainly does not tend to $0$. Conclusion: $\sum a_n$ diverges.