test the series for convergence or divergence
the series from n=0 to infinity of
(x^2+1)/(x^3+1)
I said that due to the limit comparison test this converges at 1
<<the series from n=0 to infinity of
(x^2+1)/(x^3+1)>>
Where did the x come from? Shouldn't he variable be n?
For large n, it approaches the sum of 1/n with n going to infinity, which does NOT converge. You can prove that with the "integral test". It behaves like log n as n-> infinity
To test the convergence or divergence of the series, you can use various tests, including the Limit Comparison Test. However, in this case, the limit comparison test is not the most appropriate test because it requires comparing the given series to a known series whose convergence or divergence is already known.
To test the convergence or divergence of the given series, we can use the Ratio Test. Let's apply the Ratio Test:
1. Take the limit as n approaches infinity of the absolute value of the (n+1)th term divided by the nth term:
lim (n→∞) |((x^2+1)/(x^3+1))|^n+1 / |((x^2+1)/(x^3+1))|^n
2. Simplify the expression by multiplying with the reciprocal of the term in the denominator:
lim (n→∞) |((x^2+1)/(x^3+1))|^n * |((x^3+1)/(x^2+1))| / 1
3. Simplify the expression further by canceling out the common factors:
lim (n→∞) |x^2+1|^n * |x^3+1| / |x^3+1|^n * |x^2+1|
4. Reduce the expression to:
lim (n→∞) |x^2+1| / |x^3+1|
Now, we need to analyze the limit based on the value of x:
- If |x| > 1, then lim (n→∞) |x^2+1| / |x^3+1| = |x^2+1| / |x^3+1|. As n approaches infinity, the denominator |x^3+1|^n will become exceedingly larger than the numerator |x^2+1|^n. Therefore, the series will converge.
- If |x| = 1, then lim (n→∞) |x^2+1| / |x^3+1| = 1 / 1 = 1. In this case, the result will be 1, which is inconclusive.
- If |x| < 1, then lim (n→∞) |x^2+1| / |x^3+1| = |x^2+1| / |x^3+1|. Similar to the |x| > 1 case, the denominator will become exceedingly larger than the numerator, and the series will converge.
Therefore, the series converges for both |x| > 1 and |x| < 1, and is inconclusive for |x| = 1.
Hence, your statement that the series converges at x = 1 is incorrect. The series actually converges for |x| > 1 and |x| < 1.