I am working on the following assignment about series: $$ a_n \, \, is \,\, a \,\, convergent \,\, progression \,\, and \\ \sum_1^\infty (a_n+1) = lim_{x\to\infty} {x^2-x+1\over x^2-cosx}$$ Determine if the sum converges and $$lim_{x\to\infty} a_n$$

What I have done so far: I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition). I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit: $$lim_{x\to\infty} {x^2-x+1\over x^2-cosx} = 1 $$
Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean: $$\sum_1^\infty a_n+1 = \sum_1^\infty a_n+ \sum_1^\infty 1 $$

I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?

Lots of thanks in advance!

  • $\begingroup$ See my answer below. $\endgroup$ – John Omielan Dec 27 '18 at 1:54
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    $\begingroup$ Perhaps the progression $\{a_n\}$ converges to $-1$ : $$\lim\limits_{n\to\infty}a_n=-1$$ $\endgroup$ – rsadhvika Dec 27 '18 at 1:59
  • $\begingroup$ $$\sum_1^\infty( -1+1)~~~ ?=~~~ \sum_1^\infty -1+ \sum_1^\infty 1$$ $\endgroup$ – rsadhvika Dec 27 '18 at 2:00
  • $\begingroup$ @rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found? $\endgroup$ – Artem Dec 27 '18 at 2:21
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    $\begingroup$ You knew $$\sum_1^\infty (a_n+1 )= 1$$ $\endgroup$ – rsadhvika Dec 27 '18 at 2:30

I believe the expression

$$\sum_{1}^{\infty} a_n + 1$$


$$\left(\sum_{1}^{\infty} a_n \right) + 1$$

If it meant as you suggest, the expression would instead most likely be written as

$$\sum_{1}^{\infty} \left( a_n + 1 \right)$$

This would avoid potential ambiguity.

  • $\begingroup$ Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets! $\endgroup$ – Artem Dec 27 '18 at 2:10
  • $\begingroup$ @Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity. $\endgroup$ – John Omielan Dec 27 '18 at 2:13
  • $\begingroup$ @Artem Well, the sum converges if it is equal to 1... $\endgroup$ – Osvaldo Paniccia Dec 27 '18 at 2:27

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