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We can now provide the proof of the ratio test. Recall the ratio test:
The Ratio Test 

To apply the ratio test to a given infinite series
∑ k = 1 ∞ a k , {\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k},\end{aligned}}} we evaluate the limit lim k → ∞  a k + 1 a k  = L . {\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Big }{\frac {a_{k+1}}{a_{k}}}{\Big }=L.\end{aligned}}} There are three possibilities:

Proof
Our proof will be in two parts:
 Proof of 1 (if L < 1, then the series converges)
 Proof of 2 (if L > 1, then the series diverges)
Proof of 1 (if L < 1, then the series converges)
Our aim here is to compare the given series
∑ k = 1 ∞ a k {\displaystyle \sum _{k=1}^{\infty }a_{k}}
with a convergent geometric series (we will be using a comparison test).
In this first case, L is less than 1, so we may choose any number r such that L < r < 1. Since
lim n → ∞  a n + 1 a n  = L , L < r {\displaystyle \lim _{n\rightarrow \infty }{\Big }{\frac {a_{n+1}}{a_{n}}}{\Big }=L,\quad L<r}
the ratio a_{n+1}/a_{n} will eventually be less than r. In other words, there exists an integer N such that
 a n + 1 a n  < r , w h e n e v e r n ≥ N {\displaystyle {\Big }{\frac {a_{n+1}}{a_{n}}}{\Big }<r,\quad \mathrm {whenever} \ n\geq N}
This follows from the formal definition of limit, which not all calculus students have covered. Those students who have not covered a formal definition of limit may wish to consult a calculus textbook on this (any of the textbooks listed on the Recommended Resources page cover it).
We can rearranged our expression to
 a n + 1  <  a n  r , w h e n e v e r n ≥ N {\displaystyle a_{n+1}<a_{n}r,\quad \mathrm {whenever} \ n\geq N}
If we let n {\displaystyle n} equal N, N + 1, N + 2 in the previous equation we obtain
 a N + 1  <  a N  r  a N + 2  <  a N + 1  r <  a N  r 2  a N + 3  <  a N + 2  r <  a N  r 3 {\displaystyle {\begin{aligned}a_{N+1}&<a_{N}r\\a_{N+2}&<a_{N+1}r<a_{N}r^{2}\\a_{N+3}&<a_{N+2}r<a_{N}r^{3}\\\end{aligned}}}
and, in general,
 a N + k  <  a N  r k , w h e n e v e r k ≥ 1 {\displaystyle a_{N+k}<a_{N}r^{k},\quad \mathrm {whenever} \ k\geq 1}
Now the series
∑ k = 1 ∞  a N  r k =  a N  r +  a N  r 2 +  a N  r 3 + … {\displaystyle \sum _{k=1}^{\infty }a_{N}r^{k}=a_{N}r+a_{N}r^{2}+a_{N}r^{3}+\ldots }
is convergent because it is a geometric series whose common ratio r {\displaystyle r} is known to satisfy 0 < r < 1. By the Comparison Test, the series
∑ n = N + 1 ∞  a n  =  a N + 1  +  a N + 2  +  a N + 3  + … {\displaystyle \sum _{n=N+1}^{\infty }a_{n}=a_{N+1}+a_{N+2}+a_{N+3}+\ldots }
is convergent, and so our given series
∑ k = 1 ∞ a k {\displaystyle \sum _{k=1}^{\infty }a_{k}}
is also convergent (adding a finite number of finite terms to a convergent series will create another convergent series).
Therefore, our series is absolutely convergent (and therefore convergent).
Proof of 2 (if L > 1, then the series diverges)
Here the Divergence Test implies that the given series diverges. Indeed, if
 a n + 1 a n  → L > 1 {\displaystyle {\Big }{\frac {a_{n+1}}{a_{n}}}{\Big }\rightarrow L>1}
then the ratio
 a n + 1 a n  {\displaystyle {\Big }{\frac {a_{n+1}}{a_{n}}}{\Big }}
will eventually be greater than 1; that is, there exists an integer N such that
 a n + 1 a n  > 1 , f o r n ≥ N {\displaystyle {\Big }{\frac {a_{n+1}}{a_{n}}}{\Big }>1,\quad \mathrm {for} \ n\geq N}
For this same N , {\displaystyle N,} chaining these inequalities together shows that  a N + k  >  a N  > 0 {\displaystyle a_{N+k}>a_{N}>0} for all k ≥ 1 {\displaystyle k\geq 1} . This clearly implies that the sequence { a n } {\displaystyle \{a_{n}\}} cannot converge to 0. Therefore, the given series diverges by the Divergence Test.
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There is a typo in the proof of 1. The R.H.S of equality involving the series \sum_{k=1}^\infty a_Nr^k is wrong. There should be no a_{N+1}, a_{N+2} terms on the R.H.S.
I think there is an error in the equation just above the paragraph that starts “is convergent because it is a geometric…”. To rhs of the equation should have a common factor a_N not factors a_{N+k}