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.