Numerical series

[glossy birch]

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[minor needle]

use a contradiction

[lime egret]

or contrapositive

[merry star]

i tried to build a proof by contraposition by using the definition of the limit but i don't think that it's giving any result

[lime egret]

but make sure you formulate correctly that the limit is not 0

[minor needle]

$\frac{V_n}{n} \rightarrow 0 \iff (\forall \epsilon >0)(\exists k)(\forall m > k) V_m \leq \epsilon m$

check the contrapositive

[vapid pasture]

Coffey

[minor needle]

suppose it has some non zero positive limit L

(it can't be negative obviously)

$¬\left(\frac{V_n}{n} \rightarrow 0 \right) \implies (\exists \epsilon >0)(\forall k)(\exists m > k) V_m > \epsilon m$

[vapid pasture]

Coffey

[minor needle]

what can we see from this?

Also another thing to note is, we can only ever have finite number of elements in a summation that are bigger than some number (positive sequence sum)

otherwise by Archimedian property leads to a contradiction

Ill be back after my chemistry class ends can't really put any thought to it rn

[lime egret]

could you show us what you figured out so far? @merry star

[merry star]

that's what i tried to think about so far

[minor needle]

fix an epsilon and proceed

[lime egret]

well by contrapositive, there exists a subsequence of V_n/ n that is larger than some positive number

[minor needle]

you might wanna look at V_m as elements of form 1/u_n bigger than 1/m

[lime egret]

think about what V_n means and how that translates to the elements 1/u_n

if these 1/u_n are larger than some positive number infinitely often, then a series of them surely can't converge

[minor needle]

and how we get to what aL said is by the fact that for the sum to converge the sequence describing "what we sum" must converge to 0 :3

you could also go about showing that the partial sums are arbitrarily large for large enough upper index

(positive terms are being summed upon so it's straightforward)