# Why the series is divergent based on the Preliminary test

agnimusayoti
Homework Statement:
Use the preliminary test to decide whether the following series are divergent or require further testing:
1. ##\sum_1^\infty {\frac {(-1)^{n+1}n^2}{n^2+1}}##
2. ##\sum_1^\infty {\frac {(-1)^n n^2}{(n+1)^2}}##
3. ##\sum_1^\infty {\frac {(-1)^n n}{\sqrt{n^3+1}}}##
Relevant Equations:
Preliminary test:
if ##\lim_{n\to \infty} a_n \neq 0## or ##\lim_{n\to \infty} a_n## does not exist then the series is divergent. If ##\lim_{n\to \infty} a_n = 0## then the series need further testing.
Interestingly, If I neglect the ##(-1)^n## or ##(-1)^{n+1}## then apply preliminary test, I could find the limit. Whether the limit is not equal to zero, as in series number 1 and 2, then I can conclude the series is divergent. But, if the limit is equal to zero, as in series number 3, then I can conclude the series need further testing.

I try to use the correct preliminary test, that including the ##(-1)^n## factor, then I try to find the limit as n approaching infinity. If I expand those 3 series, the sign is oscillating. Maybe, I can conclude that the limit of the series doesn't exist (or the sequences do not tend to some finite number). So the series is divergent according to preliminary test. But, surprisingly, series number 3 is oscillating too, and tends to -1 or 1 (depends on odd/even terms); yet it need further testing (which fit with explanation in the 1st paragraph).

At this point, I just realize that I don't have sufficient skill to apply preliminary test in the alternating series. Could you please guys, help me? Thanks a lot.

agnimusayoti

Homework Helper
Gold Member
Look again at whether series 3 tends to 1 or -1 as you say.

Ah, if I am not mistaken you also say two opposite things in the same post, first you say it tends to 0, later that it alternates between 1 and -1.

Homework Statement:: Use the preliminary test to decide whether the following series are divergent or require further testing:
...
3. ##\sum_1^\infty {\frac {(-1)^n n}{\sqrt{n^3+1}}}##
Relevant Equations:: , if the limit is equal to zero, as in series number 3, then I can conclude the series need further testing.

surprisingly, series number 3 is oscillating too, and tends to -1 or 1 (depends on odd/even terms); yet it need further testing (which fit with explanation in the 1st paragraph).

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