How can I prove lim (x_n)/n = k?

Question

Let x_n be a sequence of real numbers such that lim (x_n - x_(n-1)) = k. Show that lim (x_n)/n = k.

If lim (x_n)/n = k, then is it true that lim (x_n - x_(n-1)) = k?

Thank you for any help

Answer 1

Hi Tania,
The other solution you received I think is similar to mine except I went the definition of sequence convergence route, but it seems to end the same way (telescoping.)
Let epsilon (e, for us) > 0 be given, then there exists an N > 0 such that | x_n - x_n-1 - k| < e
we then have k - e < x_n - x_n-1 < k + e
and x_n < k + e + x_n-1 Similar inequality on left side. Then x_n+1 etc. so that x_n+j< (j +1)(k +e) + x_n-1
Divide by j: (x_n+j)/j < [(j+1)/j)](k +e) + x_n-1
Now as j goes to infinity the right side goes to k + e since
x_n-1 is a fixed no. divided by j which is going to infinity.

I haven't thought about the second question but I notice the other chap has a counter example, so I'll pass on that part.
Tania if you're taking a course in advanced calc. or real analysis, I'd be interested in seeing other problems.
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Answer 2

For every n>=2, x_n = x_1 + (x_2 - x1) ....+ (x_n - x_(n-1)). Observe this sum telescopes.
So, (x_n)/n = (x_1)/n + ((x_2 - x1) ....+ (x_n - x_(n-1)))/n = x_1/n + ((n-1)/(n) ((x_2 - x1) ....+ (x_n - x_(n-1))/(n-1).

lim x_1/n = 0
lim (n-1)/n = 1
((x_2 - x1) ....+ (x_n - x_(n-1))/(n-1) is the sequence of arithmetic means of (x_n - x_n -1), n=2, 3, 4. Since this last sequence converges to k, so does the sequence of it's arithmetic means.

So, lim (x_n)/n = 0 + 1* k = k, as desired.

The converse is not true. Put x_n = k*n + sin(n). Then, lim (x_n)/n = k, because sin(n) is bounded. But x_n - x_(n-1) = k + sin(n) - sin(n-1), so that it doesn't have a limit because sin(n) - sin(n-1) diverges. Actually, sin(n) - sin(n-1) = 2 sin(1) cos(n -1/2). Since this is a periodic sequence with period 2pi, irrational, it's dense in [-1, 1], so diverges.
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