For $n \in \mathbb{N}$ let $c_{n}$ be defined by
$$c_{n}=\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n$$ We have to prove that $c_{n}$ is a decreasing sequence of positive numbers.
I've already shown the first part, that it is a decreasing sequence by considering the difference of
$$c_{n+1}-c_{n} = \ln \left(1- \frac{1}{n+1} \right) +\frac{1}{n+1} $$ and then using the expansion of $\ln (1-x)$ for $-1\leq x \leq 1$.
But I'm having some trouble in showing the second part that all terms in the sequence are positive. I tried using first form of induction but but stuck in the inductive step, can somebody please suggest explain that to me? or better suggest some other way to prove that part?
Any sort of welcome as log as it leads to the solution, thanks in advance.
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Martin Sleziak
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Shreya
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3Hint. $1+\frac{1}{2}+...+\frac{1}{n}$ is Riemann upper sum of $\int_1^{n+1} \frac{1}{x}d x$. – Lukas Betz May 11 '15 at 11:12
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Have a look at the picture in this answer or this answer. You can probably find several other similar posts on this site. – Martin Sleziak May 11 '15 at 11:55
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See also http://math.stackexchange.com/questions/156326/showing-inequality-for-harmonic-series, http://math.stackexchange.com/questions/858699/how-to-prove-that-for-n-in-mathbbn-we-have-sum-k-2n-frac1k-leq and http://math.stackexchange.com/questions/701050/showing-that-sum-i-1n-frac1i-geq-logn – Martin Sleziak May 11 '15 at 11:57
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oh, yeah I got it, thanks. – Shreya May 11 '15 at 11:58
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You know that $\ln(x) = \int_1^x \frac{1}{t} dt$, right? The expression (except for the last term) is an upper sum (using a uniform partition, each partition having length 1,l i.e, with the partition points being $1, \ldots, n$) for the integral that defines $\ln(n)$, and hence is greater than or equal to $\ln(n)$.
John Hughes
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Hint: Write $\ln n$ as the integral of $1/x$ and bound each term of the Riemann sum.