No n Powers

If some of the terms are negative, then look at the series of absolute values instead. If that converges, then the original will as well, because of the Absolute Convergence Test. Try to get rid of negative exponents, and treat them as fractions instead. If you have fractions inside of fractions, try finding common denominators and reduce to just one fraction.

As an example,

= 6/((n^2+n)*n^(1/2)) <= 6 * 1/(n^2)

Then use the Comparison Test, comparing with the p-series above:

the sum over n
from 1 to infinity of 1/(n^2) converges, so the sum over n from 1 to infinity
of (3n^-2)/((n+1)(2n)^(-1)*n^(1/2)) converges

If there aren't any transcendental functions (like the natural logarithm, the tangent function, and so forth) in the term, do the following: find the largest power of n contained in the a[n] term. If the term is a fraction, find the largest power of n in both the numerator and the denominator, and subtract the largest power in the denominator from the largest power in the numerator to get the largest power of n in the entire term. If two n terms are multiplied, add the powers. If a group of terms is inside, say, a cube root, divide all powers inside the cube root by three.

The resulting highest power should be negative. If not, then find the limit of the terms: it probably won't be zero. If it is negative, then try the Comparison Test or the Limit Comparison Test with b[n] = 1/n^p, where p is the power from above. If that doesn't work, and the term looks like something you can integrate, try the Integral Test.

For instance,


has a power of 3 in the numerator, and a power of 2+2/3 in the denominator, so the whole fraction should compare favorably with n^(1/3). Use the Limit Comparison Test, and divide the above term by n^(1/3) to get


Then, in order to find the limit as n goes to infinity, divide both top and bottom by n^3 to get

(1+n^(-2.5))/(2*(3n^-1 +2)^(1/3))

and then the limit is found to be 2^(-4/3), a positive number. Since n^(1/3) diverges as a series (see p-series), the original series also diverges.

If you have an "exp" function, or a hyperbolic trig function, then write out the functions in terms of e, and you'll see that you do have an n in a power. Try those tests instead.

If you have a logarithm, then try treating it as an extremely small power of n. In fact, for any n^k, with k positive, ln n < n^k for large enough n. So,

ln(n/(n^2)) < (n^(1/2))/(n^2) =

which is a converging p-series, so the original series converges as well.

If you have a trigonometric function, check to see if you can find a pattern to the results; this is most likely if you have pi inside the trig function. If you just have a sine or a cosine function, try treating those as if they were constants; that might work, especially with a comparison test (for example, |sin x|<=1).

For example, the sum of sin(pi/2*n) is really 1 + 0 + -1 + 0 + 1 + 0 + -1 + 0 + ... Its sequence of partial sums has no limit, so the series does not converge. The sum of |sin n|/n^2 has terms smaller than 1/n^2, which converges, so the sum of |sin n|/n^2 also converges.

Copyright © 1996 Department of Mathematics, Oregon State University

If you have questions or comments, don't hestitate to contact us.