How I decided to show it is by Induction, and I don't really like what I have, so could you guys please tell me if my proof valid. Thanks.

Aim of the proof: Show that $$a_n>a_{n+1}$$ for all $n\geq1$.

Base case: $$a_1-a_2=\log(2)-\frac{1}{2}>0.$$ Hence base case holds. Assume, for some $k>1$, that the result holds, that is:$$a_k>a_{k+1}$$ Hence show that $$a_{k+1}>a_{k+2}.$$ Now we can rewrite $a_{k+1}>a_{k+2}$ as, $$\left(\sum_{k=1}^{k+1}\frac{1}{k+1}\right)-\log(k+1)>\left(\sum_{k=1}^{k+1} \frac{1}{k+1}\right)+\frac{1}{k+2}-\log(k+2),$$ Which is equivealnt of saying: $$\log\left(\frac{k+2}{k+1}\right)>\frac{1}{k+2}$$

For our problem $k\geq1$ and $\log(\frac{3}{2})>\frac{1}{3}.$

My last step is saying that as $n$ tends infinity, the limit of LHS of our last inequality is infinite and RHS is 0. Hence the result holds.

What I don't like is the fact that I don't use the inductive hypothesis at all. (Sorry for selling and grammar errors, English is not my native language)

  • $\begingroup$ This does not answer your question, but do you know that $\lim_n a_n =\gamma$, the Euler-Mascheroni constant? You might want to google it, you will for sure find something about the non decreasing behaviour of $a_n$. $\endgroup$ – Paolo Intuito Nov 23 '17 at 21:53
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    $\begingroup$ A fundamental step you make in your proof is incorrect: $\lim_{k \rightarrow \infty} (\frac{k+2}{k+1}) = 0$, not infinity, as you claim. (In fact, if this step were correct, then this alone would be a proof - the whole induction set up would be unnecessary.) $\endgroup$ – John Don Nov 23 '17 at 21:59
  • $\begingroup$ Oh crap, yeah, the fraction gets closer to 1 eqch time. Thank you for noticing this John. How would I go about proving it then?, and yes, G.S. I know about that fact but couldnt find anything on the question I am trying to solve $\endgroup$ – Scavenger23 Nov 23 '17 at 22:02

You wish to show the inequality,

$$\frac{1}{n+1}-\ln(n+1)<-\ln n$$ or equivalently, $$\frac{1}{n+1}<\ln (1+\frac{1}{n})$$

Now we have $$x-\frac{x^2}{2}<\ln(1+x)$$ so just check that $$\frac{1}{n+1}<\frac{1}{n}-\frac{1}{2n^2}$$

Which is indeed the case as it is not hard to show.

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  • $\begingroup$ How do you know that your second to last line is true? I feel like I am missing something basic. $\endgroup$ – Scavenger23 Nov 23 '17 at 22:26
  • $\begingroup$ Either, you note that the next term in the power series is positive, or you could use the mean value theorem. $\endgroup$ – Rene Schipperus Nov 23 '17 at 22:27
  • $\begingroup$ Or more straightforward, move all the terms to the right, and differentiate, and show the derivative is positive, and thus the function is increasing. $\endgroup$ – Rene Schipperus Nov 23 '17 at 22:33
  • $\begingroup$ I see now, didnt realise that you tailor expanded the log. Thanks for the help. $\endgroup$ – Scavenger23 Nov 23 '17 at 22:35

I like Rene Schipperus' method a lot - it is very simple. Here is an commonly used alternative:

Note that, the function $\frac1x$ is strictly decreasing on $[1, \infty)$, so, for all $n \in \mathbb{N}^{\, \ge 1}$,

\begin{align} \ln (n+1) - \ln n &= \int_{n}^{n+1} \frac1x \, dx\\ &> \int_n^{n+1} \frac1{n+1} \, dx\\ &= \frac1{n+1} \end{align}

As you have already noted, the solution follows.

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  • $\begingroup$ very good work. So essentially the method I wanted to use was correct, ? The question in the book never told me how to prove it. Induction seemed the way forward for me. $\endgroup$ – Scavenger23 Nov 23 '17 at 22:43
  • $\begingroup$ @KuderaSebastian That seems reasonable... I suppose that the thing to notice was that, at some point, (e.g. in your inductive step) you would have to relate $\log(n)$ to some function of integers somehow (as this is what you have on the LHS). One way is to use the power series expansion (as Rene has done), and the other is to notice that the derivative of $\log(x)$ is in fact $\frac1x$ (which both I and Peter have used in our answers). $\endgroup$ – John Don Nov 23 '17 at 22:53
  • $\begingroup$ Jon.Nice answer, short and clear. $\endgroup$ – Peter Szilas Nov 24 '17 at 9:06

Show that $ a_n -a_{n+1} >0.$

$\Delta_n: = a_n - a_{n+1} = $

$-\dfrac{1}{n+1} + \log(n+1) -\log(n).$

Mean Value Theorem:

$f(x) := \log(x)$ then:

$\dfrac{f(x+1)-f(x)}{1} = f'(t), $ $x \lt t \lt (1+x).$

$\dfrac{\log(n+1) -\log(n)}{1} = \dfrac{1}{t}$, $n\lt t\lt (n+1).$

$\Delta_n = -\dfrac{1}{n+1} + \dfrac{1}{t}$ , where $t \lt (n+1).$

Hence $\Delta_n >0.$

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  • $\begingroup$ NIce... my answer is essentially this, but integrated, in a sense. $\endgroup$ – John Don Nov 23 '17 at 22:49

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