# prove increasing/decreasing sequence

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).

• Something is wrong with these statements. I don't think they are true in their current form. – peter.petrov Mar 20 '15 at 12:04
• I proved formally and rigorously that 1) is true for any k. But I am not so sure about 2) yet, seems the behavior of that sequence depends on the value of k. I am studying it with the help of a small computer program I wrote. – peter.petrov Mar 20 '15 at 20:14
• Now I don't think 2) is true for all k. See also: math.stackexchange.com/questions/1198873/… What do you think about that? Many thanks in advance (I spent 2 days on these two sequences trying to prove things some of which were not true). – peter.petrov Mar 20 '15 at 20:38

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).