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Convergence of series

For we define and . For which does the series

converge?

Solution

We may assume that is not the zero vector and otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.

(a) If the above condition is satisfied then, without loss of generality, let be the component of maximal absolute value and let . Hence , as

and the given series is convergent because .

(b) If the above condition is not satisfied, then there are at least components of maximal absolute value and therefore

and the given series is not convergent because .

Inequality

Let be endowed with the usual product and the usual norm. If then we define . Prove that

Huygen’s Inequality

Let . Prove that

Solution

Let . is twice differentiable with

Hence is convex. The tangent at has equation . The result follows.

Sigma divisor sum

Let denote the divisor function. Prove that

Solution

We have successively:

Derivative at 0

Let

Evaluate .

Solution

The domain of is . Let us consider the logarithmic of

and differentiate it; Hence,

For we have

Thus, since .