Finding a harmonic number closest to a given large integer So, today in my real analysis course, a theorem was given:
$$\sum a_n \text{ converges }\iff \{s_n\}_{n=1}^{\infty} \text{ is bounded,}$$
where $s_n$ is the $n$-th partial sum of $\sum a_n.$ Now, of course this theorem makes perfect sense to me, but I began to wonder if there is a way to use this and only this to prove the divergence of the harmonic series. So I have been led to two questions...
a) Given $n \in \mathbb{N},$ is there a way to find the $m$-th harmonic number closest (either above or below) $n$? 
b) Given $M \in \mathbb{R},$ how can one show that there exists $N \in \mathbb{N}$ so that $\sum_{n=1}^N \frac{1}{n} > M,$ with only the above theorem and perhaps theorems primitive to it?
Thanks a lot! This isn't a homework problem, just a consideration of mine before falling asleep -- I can't exactly try to work on it myself with class in 7 hours ):
 A: To answer (a): there is an asymptotic series for the harmonic numbers, which begins
$$ H(n) \approx \log n + \gamma $$
where $\gamma \approx 0.5772156649$ is the Euler-Mascheroni constant.
If you try to invert this you get that the harmonic number which is approximately equal to $x$ should have index $\exp(x-\gamma)$.  So if you round this up, letting $m(n) = \lceil \exp(x-\gamma) \rceil$, you get a reasonable first guess of the $m$ in question 1. For example, $\exp(4-\gamma) \approx 30.655$, $H_{30} \approx 3.995$ and $H_{31} \approx 4.027$.
But of course $H_n$ isn't exactly given by the formula I gave. In particular that formula is an underestimate, so  it should overestimate $m(n)$.  In fact this sequence is A002387 in the Encyclopedia of Integer Sequences, where it is conjectured that $m(n) = \lfloor \exp(n-\gamma)+1/2 \rfloor$. This is either equal to or one less than my crude estimate.
A: Ok, correct me if I've understood the question incorrectly, but:


*

*terms in harmonic series are positive hence $s_n\uparrow$ with $n\uparrow$.

*you can easily verify that $s_{2^k}-s_{2^{k-1}}\geq \frac12$: just estimate each $a_n\geq\frac{1}{2^k}$ for $2^{k-1}+1\leq n\leq 2^k$.  So the set $\{s_n\}_{n\geq 1}$ is unbounded and the harmonic series does not converge.

*Hence the convergence is monotonic from (1.) for any $M>1$ there is such $n:s_n\leq M\leq s_{n+1}$. Indeed, if there is such $M^*$ for which our assumption fails, we obtain the boundness of $\{s_n\}$ which we have already disproved. 

*Clearly, if $s_n\leq M \leq s_{n+1}$ then 
$$
|s_{n-1}-M| = M- s_{n-1} = M -s_n+\frac1n = |M-s_n|+\frac1n>|M-s_n|
$$
so $s_n$ is the best lower bound for $M$. In the same way you show that $s_{n+1}$ is the best upper bound for $M$, so we have only two candidates for the closest harmonic number.

*Computationally the way how you find such $n$ for a given $M$ is either by simple iterations, starting from $n=1$ until you will overcome $M$, or by using that 
$$
\log(n+1)\leq s_{n}\leq 1+\log{n}.
$$
This formula can be obtained by using the fact that 
$$
\log (n+1)-\log n=\int\limits_n^{n+1}\frac{dt}{t}\in\left(\frac{1}{n+1},\frac1n\right).
$$
As a result you obtain for $n:s_n\leq M\leq s_{n+1}$ that $\mathrm e^{M-1}\leq n\leq \mathrm e^M$.
