Ytukyg

Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$

Must there be an index k for which $q_k > 5^{50}$?

It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$

But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$

Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.

Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.

Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.

Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.

If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.

Therefore the series is not bounded and will eventually reach any arbitrarily large number.