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I am studying this sequence with some help from a small computer program which I wrote.

$(1 + \frac{k}{n})^{(n+1)}$

I am studying it for different integer values of k (positive and negative).

I have so far built this theory:

1) For k < 0 the sequence is increasing after certain value of n
2) For k = 1,2 the sequence is decreasing after certain value of n
3) For k >= 3 the sequence is increasing after certain value of n

I know that its limit is $e^k$ but I want to know about its monotonicity.

Can anyone prove or reject this hypothesis? I am looking for some simple solution based only on math induction and on basic facts about sequences. Assume that I know all basic theorems about sequences, monotonicity, convergence, etc. and that I have proved the following somewhat related statement.

For all integers k the sequence $(1 + \frac{k}{n})^{n}$ is increasing
(at least for all sufficiently large n) and its limit is $e^k$
.

I also know Bernoulli's inequality (which seems useful for such problems).

Earlier today I built another hypothesis which now seems wrong to me.

prove increasing/decreasing sequence

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  • $\begingroup$ For the cases $k=1,2$ is decreasing not increasing. $\endgroup$ – science Mar 20 '15 at 23:15
  • $\begingroup$ @science Yes, that's what I meant. I have swapped them by mistake. Fixed. By the way I am interested in both positive and negative values of k. $\endgroup$ – peter.petrov Mar 20 '15 at 23:19
  • $\begingroup$ What do you mean? That for some k (actually for all k but 1 and 2), it is neither decreasing nor increasing, is that what you mean? Could you elaborate a bit? But I think I am seeing an obvious pattern with a computer as the sequence goes closer to its limit (even for k>=3 and for k<0). $\endgroup$ – peter.petrov Mar 20 '15 at 23:53
  • $\begingroup$ The case $k=-1$ is increasing not decreasing. However you are saying it is decreasing! $\endgroup$ – science Mar 21 '15 at 0:03
  • $\begingroup$ Right, sorry, I've fixed my theory again. So that's quite interesting. But is it true and how it can be proved? I will think some more myself. Sounds like a nice puzzle. $\endgroup$ – peter.petrov Mar 21 '15 at 0:27
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We define $f(x)=(1+k/x)^{x+1},$ and want to show this is eventually decreasing if $k=1,2$ while it is eventually increasing if $k\ge 3.$ If by $\exp(a)$ we mean $e^a,$ then we have $f(x)=exp(g(x))$ where $$g(x)=(x+1)\ln(1+k/x).$$ Note that since $\exp(u)$ is strictly increasing, we know that $f$ is increasing/decreasing iff $g$ is so.

The derivative of $g(x)$ is now $$g'(x)=\ln(1+k/x)-\frac{k(x+1)}{x(x+k)},\tag{1}$$ after simplifying it.

Now we can apply the series for $\ln(1+t)=t-t^2/2+t^3/3 \cdots$ to this, and note since we're only interested in eventually large $x$ that the value $k/x$ is eventually less than $1$ so that the log series will converge when $t=k/x.$ Furthermore note that then the log series is an alternating series whose terms have strictly monotone decreasing absolute values. This means that "tail ends" of the log series have definite sign, specifically a tail starting with a positive term has a positive sum, and a tail starting at a negagive term has a negative sum.

Define $h(k)$ as the term $k(x+1)/[x(x+k)]$ which is the fraction subtracted from the log term on the right side of $(1).$ We first assume that $k\ge 3$ and we take the first two terms of the log series, namely $k/x-(1/2)(k/x)^2,$ and subtract off $h(k)$ which then gives $$\frac{k[(k-2)x-k^2]}{2x^2(x+k)}.$$ Since $k \ge 3$ this is eventually positive. Now putting the rest of the log series back we're beginning at a positive term $+(1/3)(k/x)^3,$ so that the total sum remains positive ande we have shown eventually $g'(x)>0$ in case $k \ge 3.$

In the two cases for $k=1,2$ we need to use the first three terms of the log series, and then subtract $h(k)$ as above, this time wanting the result to be eventually negative, because now the first term in the tail is the negative value $-(1/4)(k/x)^4$ making the tail cause the total series to give a negative sum and so get $g'(x)<0$ for large $x.$ What we find is that for $k=1$ our first three terms with $h$ subtracted is $(2-3x)/6x^3,$ while in case $k=2$ this is $-4(x-4)/(3x^3(x+3)),$ in each case eventually negative as desired in order to show $g'(x)<0$ for large $x.$

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