Are these two statements true? If so, how does one prove them?
1) For each integer k (positive or negative), the sequence
$a_n = (1 + k/n) ^ n$ (1)
is increasing (at least after a certain number n).
2) For each integer k (positive or negative) the sequence
$a_n = (1 + k/n) ^{n+1} $ (2)
is decreasing (at least after a certain number n).
Note that I have no problem proving that the two sequences are convergent and to find their limits but I have really hard time proving formally that these are monotonic (after certain n).
In fact for statement (1) seems one can prove it this way.
Note that:
$(1 + \frac{k+1}{n} )^n = \frac{(1 + \frac{k}{n+1})^{n+1}(1 + \frac{1}{n})^{n}}{1+\frac{k}{n+1}}$ (3)
Now apply induction on k. Based on (3), the (k+1)-sequence ${a_n}$ must be increasing because on the right hand side we have: the increasing k-sequence ${a_{n+1}}$ multiplied by the increasing 1-sequence ${a_n}$, and then divided by a decreasing sequence. So the left hand side must be increasing too.
Is this proof correct? Actually I think it only works for k>0 otherwise the sequence in the divisor (1 + k/(n+1)) is not decreasing but is increasing. Also, what is the proof for (2)? Should be something analogous, I guess, like the above proof for (1).
Best Answer
Consider \begin{align*} \log(a_n) &= n\log\left(1 + \frac{k}{n}\right) \\ &= n\left(\frac{k}{n} - \frac{(\frac{k}{n})^2}{2} \mp \ldots \right)\\ & = \left(k - \frac{k^2}{2n} + \frac{k^3}{3n^2} \mp \ldots \right) \end{align*} where the series converges absolutely for $n > k$.
Then \begin{align*} \log(a_n) - \log(a_{n-1}) &= \left(k - \frac{k^2}{2n} + \frac{k^3}{3n^2} \mp \ldots \right) - \left(k - \frac{k^2}{2(n-1)} + \frac{k^3}{3(n-1)^2} \mp \ldots \right) \\ &= k^2\left(\frac{1}{2n - 2} - \frac{1}{2n}\right) + O(k(k/n)^2) \end{align*} This is negative for $n >> k$, yielding the result. Same technique works for (2).