/sci/ - Science & Math

>so many seethy undergrads in this thread
fucking pathetic lol

>>10855853
>>/mg/ is for smart people
I would argue as much, yes.
>>/sci/ is for smart people
No, I would disagree. This is what I meant by "surroundings" if you had read my post.
>>Using certain words affects this in any sense
Yes, of course it does. If all you did was babble incoherently about mayonnaise, I would have good reason to imagine that you are not intelligent. In a similar sense, one who constantly spews derogatory language when given the opportunity most likely has little self awareness and control. This leads me to believe that they lack a propensity for concentrated study and self-improvement.
>You are a massive faggot.
I'm aware; I am indeed not a heterosexual. But is it really necessary to use such vile language?

>>10810567
I'll give you a rigorous proof if you'd like.
Denote the partial sums of the harmonic series by [math] H_n [/math]. We aim to show divergence, in fact divergence to [math] +\infty [/math]. We pass to the subsequence [math] A_k = H_{2^k} [/math]. Note that [math] A_0 = 1 [/math]. Then for [math] k > 0 [/math] we find [math] A_k = \sum_{j=1}^{2^k} \frac{1}{j} = \sum_{j=1}^{2^{k-1}}\frac{1}{j} + \sum_{j=2^{k-1}+1}^{2^{k}}\frac{1}{j} \geq A_{k-1} + \sum_{j=2^{k-1}+1}^{2^{k}}\frac{1}{2^k} = A_{k - 1} + 2^{k-1}\frac{1}{2^k} = A_{k-1} + \frac{1}{2}[/math]. Thus [math] A_k \geq A_{k - 1} + \frac{1}{2} [/math]. Summing the inequalities from [math] k = 1 [/math] to [math] N [/math], we find that [math] A_{N} \geq 1 + \frac{N}{2} [/math]. Thus, for each [math] R > 0 [/math], pick [math] N > 2R [/math] and then [math] A_N \geq 1 + \frac{N}{2} > R [/math]. This shows that [math] A_k [/math] diverges to [math] +\infty [/math].
Of course, it is a basic fact that divergence of a subsequence implies divergence of the sequence [math] H_n [/math]. Furthermore, [math] H_n [/math] is monotone, and monotone divergent series must diverge to [math] +\infty [/math]. This completes the proof.