I am given a sequence ${x_n}=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ and I have to show that it's bounded.
I tried doing this: $$1\leq 1+\frac{1}{2}+...+\frac{1}{n}\leq n$$ $$1-\ln n\leq 1+\frac{1}{2}+...+\frac{1}{n}-\ln n\leq n-\ln n$$ $$1-\ln n\leq{x_n}\leq n-\ln n$$ But I don't think I am allowed to do this since the boundaries depend on n. Is there maybe another way to solve this? and could you please help me? Thanks a lot! I can't use integrals, derivatives or limits
You have
$$x_{n+1} - x_n = \frac{1}{n+1} -\ln \left(1+ \frac{1}{n}\right)= -\frac{1}{2n^2} + O(\frac{1}{n^3})$$
As $\sum 1/n^2$ converges, you get the desired result.
Hint for an elementary solution, without integral or asymptotic approximation :
Define the sequences $(x_n)_{n \geq 1}$ and $(y_n)_{n \geq 1}$ by $$x_n = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln(n) \quad \quad \quad \text{and} \quad \quad \quad y_n = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln(n+1)$$
(under these three conditions, we say that $(x_n)$ and $(y_n)$ are adjacent sequences).
You have too many limitations "no integrals, derivatives or limits". I will solve this problem using the elementary inequality $$x-\dfrac{x^2}{2}\lt \ln(1+x)\lt x$$ for all $x\ge 1.$ In case you haven't seen it before, lets prove it applying the mean value theorem for the function $f(t)=\ln(1+t)$ on the interval $[0, x].$ According to MVT, there exists $c\in (0,x)$ such that $f(x) - f(0) = f'(c)(x-0),$ and also we have $\dfrac{1}{1 + x}\lt f'(c)=\dfrac{1}{1 + c} \lt 1$. Hence $$\dfrac{x}{1 + x} \lt \ln(1 + x) \lt x,\qquad \forall x\gt0.$$ Since $$x-\dfrac{x^2}{2}-\dfrac{x}{1 + x}=\dfrac{x^2(1-x)}{2(1+x)}\le0$$ for all $x\ge 1,$ we have the desire result. Now lets observe that \begin{align} x_n & = 1+\dfrac12+\cdots+\dfrac1n-\ln n \\ & = \left(1+\ln 1-\ln 2\right) + \left(\frac{1}{2}+\ln 2-\ln 3\right) + ... + \left(\frac{1}{n}+\ln n-\ln (n+1)\right) +\ln\left(\dfrac{n+1}{n}\right)\\ & = \sum_{k=1}^n\left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right) + \ln\left(1+\frac{1}{n}\right). \end{align}
But according to the inequality that we proved before $$0\lt\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\lt\dfrac{1}{2k^2}.$$ Since the infinite series correspond to the right most term of this last inequality is convergent $$0\lt x_n\lt \dfrac12\sum_{k=1}^{\infty}\dfrac{1}{k^2}+\max_{k\in\mathbb{N}}\ln\left(1+\frac{1}{k}\right)\lt\infty.$$ In fact, $\dfrac{\pi^2}{12}+\ln 2$ is an explicit upper bound for your sequence. Furthermore, if $$y_n = 1+\dfrac12+\cdots+\dfrac1n-\ln (n+1)$$ then $x_n=y_{n-1}+\dfrac1n$ and therefore one sequence converge if and only if so does the other. Also both of them share the same limit. Now using the same argument for $y_n$ we can get rid of $\ln 2$ term and obtain a much closer approximation to the potential limit.
