Let and be two sequences that satisfy the relationship
for all .
- Prove that if for all then for all .
- Prove that if for all and if is convergent, then
- Proof. Assume for all . From the given relation between the and we have
Now, we claim for all . To see this, consider the function
Since for all we have is increasing for all . Since we must have for all . Since by assumption we then have
But, this implies
for all
- Proof. To prove converges we use the limit comparison test with . First, since converges we know
Now, we use the given relation between the and the ,
Then, use the expansion of the exponential,
This gives us
Therefore,
Therefore, by the limit comparison test the series and either both converge or both diverge. Since converges by hypothesis, we have established the convergence of
Not necessary to do the expansion in (b). Represent b_n in terms of a_n, and apply L’Hopital rule twice.
Your proof is almost right. From the Taylor series you can deduce that the series of (e^bn-1) converges by the limit comparison test. But from part (a), we know that bn0 (because an>0, then bn>0). This shows that the series of bn converges.
Sorry, it should be bn<e^bn-1 instead of bn0
Apparently my comment didn’t post. There is a problem where you used the Taylor series in the comparison – it should be the log of the Taylor series. But more troubling to me is this: if a(n) converges, then a(n) becomes arbitrarily small. So e^a(n) —> e^b(n) because the order of magnitude of e^x is much greater then x. (x=a(n)).
So e^a(n)/e^b(n) —> 1 implying a(n) —-> b(n). So if limit b(n)/a(n) —> 1 which ≠ 0 how can this converge?
Okay, got it now. I guess a second application of Taylor was implied for the log function.
In the comparison test, b(n) should be the log of a Taylor series. Would this work? e^a(n) = a(n) + e^b(n) >e^b(n) (since a(n)>0) implying a(n)>b(n) (since b(n)>0) so b(n) converges.