Sequences for real numbers

Question

Consider the sequence {r^n / n} for r a real number.

a. Explain why the sequence is not monotonic for r < 0.

b. Show that the sequence is increasing for r ≥ e.

c. Show that the sequence is decreasing for 0 < r ≤ 1.

d. For what values of r is the sequence convergent?

e. For what values of r is the sequence divergent?

Answer 1

if r<0, terms alternate between negative and positive, not monotonic

look at function y=r^x/x

y'=[(-r^x+x*ln(r)*r^x]/x²

y'=0 when x=log_r(e), critical point

if r<e then x>1, so no guarantee from x=1 to x=2 that sequence increases (ex r=2, x=1 and x=2 have same sequence entries)

r≥e then x≤1 so guaranteed sequence increases from x=1 to x=2 and then to x=3, x=4, ...

correction convergent when last terms approach a value

L=lim rⁿ/n,,,,when n approach infinity

lnL=lim nln(r)-ln)n),,,,n large

r≥1,,,,,,ln(r)>0,,,,,,,lnL as n approaches infinity will grow and L does not approach any value and is

divergent

r<0,,,ln(r)<0,,,,lnL more and more negative values as n approaches infinity. and approaches minus infinity

convergent with L=0

Answer 2

I am assuming that this is a calculus question, not precalculus question.

As a preliminary, for any r > 0 define the function

f(x) = r^x/x

This can be rewritten as

f(x) = e^xln(r)/x

Take the derivative

f'(x) = (xln(r)e^xln(r) - e^xln(r))/x² = (xln(r) - 1)e^xln(r))/x² = (xln(r) - 1)r^x/x²

a.

For r < 0 the sequence is not monotonic because

even elements are positive, and odd elements are negative.

b.

The values of the sequence are given to us by {f(n)}.

If the function f(x) is increasing for x ≥ 1 then so is the sequence.

The function is increasing on the interval [1, ∞) if f'(x) > 0 on the interval (1, ∞).

Consider the above calculated derivative f'(x).

The factor r^x/x² > 0, so it is only the factor (xln(r) - 1) that determines the sign of f'(x).

For r ≥ e, ln(r) ≥ 1.

So for x > 1, xln(r) > 1.

That makes the f'(x) > 0, which proves that the sequence is increasing.

c.

Use the same reasoning as in b.

For 0 < r ≤ 1, ln(r) ≤ 0.

So for x > 1, xln(r) ≤ 0.

Therefore xln(r) - 1 < 0, therefore f'(x) < 0, therefore the function is decreasing,

therefore the sequence is decreasing.

Note: This raises the interesting question of what happens when 1 < r < e.

Then 0 < ln(r) < 1.

For small values of x, xln(r) < 1, but for large x, xln(r) > 1.

So the sequence starts decreasing and then increasing.

d.

We need to see which values of r give us existence of lim(rⁿ/n) as n -> ∞.

The answer is -1 ≤ r ≤ 1.

The reason is than in this interval |rⁿ| ≤ 1, while the denominator increases to ∞.

In contrast, for r > 1 the numerator rⁿ increases faster than the denominator.

(You can convince yourself by applying the L'hopital rule to the function f.)

That makes the limit is infinity.

For r < -1 the sequence in magnitude has the same values as for -r,

but it alternates between positive and negative values.

Therefore the limit is not even +∞ or -∞.

e.

As shown in d. the sequence diverges for r < -1 and r > 1.