4
$\begingroup$

Again, shattered by this question on series, I did have no clue how to begin. Sequences limits are approached through absolute values of the $n$-th term and the assumed limit being smaller than a given delta. And this for a given $N$. I don't see the link with power-series exercises, as for an example, the following:

Find a $K$, such that for all $n \geq K$ we have $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \ldots + \frac{1}{\sqrt{n}} > 1000$$

Help me with this one to, but more interesting is, how the theory on sequences applies to these problems. Or don't they?

  • 0
    the epsilon delta argument for divergence is slightly different than that for convergence, in order to prove that a sequence diverges, we say that for any epsilon, there exists a delta s.t. n>delta implies S(n) > epsilon. An easy and surefire delta for that series is is delta = epsilon^2, for the minimum delta a slightly more complex formula would be necessary. This problem above is just giving a example epsilon (1000) and asking for any specific delta.2016-08-14

5 Answers 5