The integral test

First example

  • We introduced series \(\sum a_n\) with examples where it was possible to give a closed form for the partial sums \(s_n\), and thus verify directly on the sequence \(\{s_n\}\) whether the series \(\sum a_n\) is convergent or divergent, and if it's convergent, what is its limit.
    • This is not always the case. If you can't find a closed form for the partial sums, then you need to use other ways to decide on convergence. In the next few sections, we'll look at methods do to this.
  • Example. Consider the series \(\sum_{n=1}^\infty\frac{1}{n^2}\).
    • Check out this picture
    • What you can see on it is the right endpoint approximation of \(\int_1^\infty\frac{1}{x^2}\,\mathrm dx\) with subinterval length \(\Delta x=1\).
    • You can check that since \(\Delta x=1\), the Riemann sum is exactly \(\sum_{n=2}^\infty\frac{1}{n^2}\).
    • Since we're using the right endpoint approximation, and \(f(x)\) is decreasing on \([1,\infty)\), the Riemann sum is smaller than the improper integral: \[ \sum_{n=2}^\infty\frac{1}{n^2}\le\int_1^\infty\frac{1}{x^2}\,\mathrm dx. \]
    • Therefore, since \(\int_1^\infty\frac{1}{x^2}\,\mathrm dx=\lim_{r\to\infty}\left[-\frac{1}{x}\right]_1^r=1\), in particular the improper integral is convergent, and \(f(x)\ge0\) on \([1,\infty)\), the series is convergent too.

Second example and statement of the method

  • Example. Consider the series \(\sum_{n=1}^\infty\frac{1}{\sqrt n}\).
    • Check out this picture
    • This time, we're looking at the left endpoint approximation of \(\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx\) with subinterval length \(\Delta x=1\).
    • The Riemann sum is \(\sum_{n=1}^\infty\frac{1}{\sqrt n}\).
    • Since we're using the left endpoint approximation, and \(f(x)\) is decreasing on \([1,\infty)\), the Riemann sum is larger than the improper integral: \[ \int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx\le\sum_{n=1}^\infty\frac{1}{\sqrt n}. \]
    • Since the improper integral is divergent: \(\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx=\lim_{r\to\infty}\left[2\sqrt x\right]_1^r=\infty\), and \(f(x)\ge0\) on \([1,\infty)\), the series is divergent too.
  • Integral test. Let \(f\) be a continuous, decreasing, nonnegative function on \([1,\infty)\), and let \(a_n=f(n)\). Then the series \(\sum_{n=1}^\infty a_n\) is convergent if and only if the improper integral \(\int_1^\infty f(x)\,\mathrm dx\) is convergent. In other words:
    • If \(\int_1^\infty f(x)\,\mathrm dx\) is convergent, then \(\sum_{n=1}^\infty a_n\) is convergent.
    • If \(\int_1^\infty f(x)\,\mathrm dx\) is divergent, then \(\sum_{n=1}^\infty a_n\) is divergent.
  • Exercises. 11.3: 3, 7, 13, 17, 21, 25.

Remainder estimate for the integral test

  • Let \(f(x)\) be a continuous, decreasing, nonnegative function on \([1,\infty)\), and let \(a_n=f(n)\).
    • Suppose that \(\sum_{n=1}^\infty a_n=s\).
    • We can use improper integrals to estimate partial sums \(s_n\).
    • Note that the remainder is \[ R_n=s-s_n=a_{n+1}+a_{n+2}+\dotsb \]
    • The exact same way as for the whole series, we can show that \[ \int_{n+1}^\infty f(x)\,\mathrm dx\le R_n\le\int_n^\infty f(x)\,\mathrm dx \]
    • This is called the remainder estimate for the integral test. Adding the partial sums \(s_n\) back, we get: \[ s_n+\int_{n+1}^\infty f(x)\,\mathrm dx\le s\le s_n+\int_n^\infty f(x)\,\mathrm dx \]
  • Exercises. 11.3: 36, 37.

The comparison tests

First example and statement of method

  • Example. Consider the series \(\sum_{n=1}^\infty\frac{1}{3^n+2}\).
    • Instead of directly trying to find a closed formula for the partial sums, we can notice that we have \(\frac{1}{3^n+2}<\frac{1}{3^n}\) for all \(n\ge1\).
    • This implies that \(0\le\sum_{n=1}^\infty\frac{1}{3^n+2}\le\sum_{n=1}^\infty\frac{1}{3^n}=\frac12\). In particular, the original series is convergent.
  • Comparison test. Let \(\sum a_n\) and \(\sum b_n\) be series with all terms nonnegative. Suppose that \(a_n\le b_n\) for all \(n\).
    • If \(\sum b_n\) is convergent, then \(\sum a_n\) is convergent too.
    • If \(\sum a_n\) is divergent, then \(\sum b_n\) is divergent too.
  • Exercises. 11.4: 3, 5, 9, 13, 15, 29.

The limit comparison test, estimating sums

  • We've seen in the examples that if we try to find out whether a given series is convergent, we can look for \(b^n\) with largest \(b\), or \(n^k\) with largest \(k\).
    • Then we would have to find a suitable comparison series.
    • Instead, there a more direct procedure to compare series.
    • Limit comparison test. Let \(\sum a_n\) and \(\sum b_n\) be series with positive terms. If \[ \lim_{n\to\infty}\frac{a_n}{b_n}=c \] where \(c>0\) is a finite number, then either the series both converge, or both diverge.
  • Exercises. 11.4: 17, 19, 23, 27, 31
  • Note that we can use comparisons for remainder terms too.
    • Estimates with the comparison test. Let \(\sum a_n\) and \(\sum b_n\) be series with nonnegative terms, and suppose that \(a_n\le b_n\) for \(n\ge N\).
    • Then we have \[ \sum a_n=s_N+R_N\le s_N+\sum_{n=N}^\infty b_n. \]
    • If \(\sum b_n\) is simpler to compute, then this can give estimates for \(\sum a_n\).
    • Exercises. 11.4: 33, 34.