I am trying to use the comparision test over here $Y= 1+ \frac{1}{2}+\frac{1}{3}+..$ . Show that $0<(x_n)<=(y_n)$ . Since $Y$ diverges we see that the $X$ also diverges.
$(1)$ am I using the comparision test correctly?
$(2)$is there any other crude way to prove this instead of using theorems
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If so, we obtain: $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...=$$ $$=1+(X-1)+\frac{1}{2}(X-1)+\frac{1}{2^2}(X-1)+\ldots=1+\frac{1}{1-\frac{1}{2}}(X-1)=2X-1,$$ which says that $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots$$ converges, which is wrong.
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by term-by-term comparison the given sum is greater than $ \frac 1 4 \left(1+\frac 12+\frac 1 3+\ldots\right)$.
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The series is strictly larger (at precisely each term) than the sum of the reciprocals of the primes which is known to diverge.
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Here’s another approach. By Riemann sums, $$\sum_{k=1}^{n-1}\frac1{2k-1}\geq\int_1^n\frac1{2x-1}dx=\frac12\ln(2n-1),$$ so that the sum to infinity diverges.
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$\dfrac{1}{2n-1}\gt \dfrac{1}{2n}= (1/2)\dfrac{1}{n}$.
The harmonic series $\sum \dfrac{1}{n}$ diverges.
In context:
1) The sum of the reciprocals of primes diverges as well.
https://en.m.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
The sum of reciprocals of twin primes converges.
Is the series $X =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+..$ convergent or divergent.
It is divergent, since by the comparison test $$ 1+\frac{1}{3}+\frac{1}{5}+\cdots \geq \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots $$ which is $\tfrac12$ times the famous harmonic series $$ 1+\frac{1}{2}+\frac{1}{3}+\cdots $$ which is known to diverge.