drjohnsonn
- 11
- 1
1. show that the sum of. The reciprocals of the primes is divergent. I am reposying this here under homework and deleting the inital improperly placed post
2. Theorem i use but don't prove because its assumed thw student has already lim a^1/n = 1.
The gist of the approach I took is that∑1/p = log(e^∑1/p) = log(∏e^1/p) and logx→ ∞ as x→∞.
Proof outline: let ∑1/p = s(x). (...SO I can write this easily on tablet) and note that e^s(x) diverges since e^1/p > 1 for any p and the infinite product where every term exceeds 1 is divergent. Then loge^s(x) diverges as logs as x→∞ would. Thus, since log(e^s(x)= s(x), the sum is found to be divergent
Edit: this is wrong and i finished the proof using very little ofwhati tried here so no need to respond
2. Theorem i use but don't prove because its assumed thw student has already lim a^1/n = 1.
The gist of the approach I took is that∑1/p = log(e^∑1/p) = log(∏e^1/p) and logx→ ∞ as x→∞.
Proof outline: let ∑1/p = s(x). (...SO I can write this easily on tablet) and note that e^s(x) diverges since e^1/p > 1 for any p and the infinite product where every term exceeds 1 is divergent. Then loge^s(x) diverges as logs as x→∞ would. Thus, since log(e^s(x)= s(x), the sum is found to be divergent
Homework Statement
Edit: this is wrong and i finished the proof using very little ofwhati tried here so no need to respond
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