Positive Series: Integral Test & p-series

Positive Series: Integral Test & p-series Calculus II Josh Engwer TTU 3 March 204 Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 / 8

Bad News about Summing a (Convergent) Series... Certain types of series can be summed without problem: Geometric Series Telescoping Series Unfortunately, in general, series can be hard or impossible to sum. Case in point: Euler showed in 735 that k= k 2 = + 2 2 + 3 2 + 4 2 + 5 2 + = π2 6 To this day, nobody has found the closed-form sum of Therefore, one can only determine the convergence of most series. To this end, many convergence tests will be introduced. k= k 3 Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 2 / 8

Divergence Test Theorem (Divergence Test) lim a k 0 = series a k diverges. k Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 3 / 8

Divergence Test Theorem (Divergence Test) lim a k 0 = series a k diverges. k The converse is not true in general: series a k diverges = COUNTEREXAMPLE: k= The inverse is not true in general: COUNTEREXAMPLE: lim a k 0 k diverges even though lim k k k = 0 lim a k = 0 = series a k converges k lim k k = 0 even though k diverges k= The contrapositive is true: series a k converges = lim a k = 0 k Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 4 / 8

Divergence Test Theorem (Divergence Test) lim a k 0 = series a k diverges. k PROOF: It s easier to prove the contrapositive: series a k converges = lim a k = 0 k Suppose a k converges. Then a k = L for some finite value L. = lim n = L n where S n is the n th partial sum of the the series a k = lim n = L and S k S k = a k n = lim k = lim k S k ) = lim k lim k = L L = 0 k k k k lim k = 0 k QED Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 5 / 8

Positive Series Definition ak is a positive series if each term a k 0 k. Examples of Positive Series: k 2 = + 2 + 4 + 8 + 6 + k= π k = π + π 2 + π 3 + π 4 + π 5 + k= ( ) k cos(kπ) = + + + + + k=0 Examples that are not Positive Series: ( ) k+ k = 2 + 3 4 + 5 6 + k= cos(kπ) = + + k=0 Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 6 / 8

Integral Test Theorem (Integral Test) Let function f (x) be continuous & positive for x [N, ). Moreover, suppose a k = f (k) for k = N, N +, N + 2, N + 3,... Then: Positive Series Positive Series a k converges Integral k=n a k diverges Integral k=n N N f (x) dx converges f (x) dx diverges REMARK: The Integral Test is useless for factorials (k!) because the generating curve involves the Gamma Function Γ(α) := and the Gamma Function is too complicated to work with. 0 x α e x dx, REMARK: The only hope for convergence is that f is eventually decreasing: lim x Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 7 / 8

Integral Test Theorem (Integral Test) Let function f (x) be continuous & positive for x [N, ). Moreover, suppose a k = f (k) for k = N, N +, N + 2, N + 3,... Then: Positive Series a k converges Integral k=n N Positive Series a k diverges Integral k=n N PROOF: M Recognize that a k is a Riemann Sum for k=n M The details are tedious see the textbook if interested. QED N f (x) dx converges f (x) dx diverges f (x) dx with x k =. Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 8 / 8

Useful Inequalities for the Integral Test If the resulting improper integral is hard or impossible to compute, consider applying a useful inequality leading to a simpler integral: cos x sin x π 2 < arctan x < π 2 For x R: x 2 0, x 4 0,, x 2n 0, 2 x > 0, e x > 0, x x 0 For x 0: x 0, 3 x 0, 4 x 0,, n x 0, x p 0 For x : x, 3 x, 4 x,, n x, x p For x : log 2 x 0, ln x 0, log x 0 A < B = A > B A > B = A < B A B = A B A B = A B A, M, m > 0 s.t. M > m = AM > Am and A M < A m A, x > 0 = A + x > A = A + x < A A > x > 0 = A x < A = A x > A f is positive & increasing on [A, B] AND 0 < A < B = 0 < f (A) < f (B) f is positive & decreasing on [A, B] AND 0 < A < B = f (A) > f (B) > 0 (Integral Dominance Rule) f (x) g(x) = N f (x) dx N g(x) dx Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 9 / 8

A Note about Inequality Chains Every inequality in an inequality chain must be pointing in same direction: A B C D = E F G = H implies that A H A < B C < D = E < F G = H implies that A < H A < B < C < D = E < F < G = H implies that A < H A = B C = D E F G H implies that A H A = B C = D E > F G H implies that A > H A = B > C = D > E > F > G > H implies that A > H Otherwise, the inequality chain is useless: A < B C > D = E < F A and D are related A and E are related A and F are related B and D are related B and E are related B and F are related C and F are related implies nothing on how: Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 0 / 8

Applying Inequalities for the Integral Test WORKED EXAMPLE: Test the series k=0 e k2 for convergence. Observe that for x 0: e x2 is continuous & positive, and x 2 x = x 2 x = e x2 e x (since e x is positive & increasing) Apply the Integral Test: [ dx e x dx = e x] x e x2 0 0 x=0 FTC ( = lim e x ) ( e (0)) = < x 0 e x2 dx < = 0 e x2 dx converges = k=0 e k2 converges Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 / 8

Applying Inequalities for the Integral Test 3 k5 + k + WORKED EXAMPLE: Test the series k k= for convergence. 3 x5 + x + Observe that for x : is continuous & positive, and x x 5 + x + x 5 + x x 5 and 3 x is positive and increasing. Apply the Integral Test: 3 x5 + x + dx x 3 x 5 x dx = x 5/3 x dx x dx = 3 x5 + x + dx = x 3 x5 + x + dx diverges x = k= 3 k5 + k + k diverges Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 2 / 8

Applying Inequalities for the Integral Test WORKED EXAMPLE: Test the series Observe that for x 2: (7 + 3 sin k) for convergence. k= 2 7 + 3 sin x is continuous & positive, and sin x = 3 sin x 3 = 7 + 3 sin x 7 + ( 3) = 4 Apply the Integral Test: 2 (7 + 3 sin x) dx 2 4 dx = [ ] x ( ) FTC 4x = lim 4x 4( 2) = + 8 = x= 2 x 2 (7 + 3 sin x) dx = 2 (7 + 3 sin x) dx diverges = (7 + 3 sin k) diverges k= 2 Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 3 / 8

p-series Test Theorem (p-series Test) p > = p-series p = p-series k= k= k p converges k p diverges Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 4 / 8

p-series Test Theorem (p-series Test) p > = p-series p = p-series k= k= k p converges k p diverges PROOF: Apply the Integral Test. CASE I: p =. [ ] x x dx = FTC ln x = ln( ) = = x= Harmonic Series k diverges k= Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 5 / 8

p-series Test Theorem (p-series Test) CASE II: p <. p > = p-series p = p-series k= k= k p converges k p diverges Then p < = p > = p > 0. [ ] x p x x p dx = FTC = ( ) p p x= p p = p = p-series k p diverges k= Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 6 / 8

p-series Test Theorem (p-series Test) CASE III: p >. QED p > = p-series p = p-series k= k= k p converges k p diverges Then p > = p < = p < 0. [ ] x p x x p dx = FTC = ( ) p p x= p p = 0 p = p < p-series k p converges k= Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 7 / 8

Fin Fin. Josh Engwer (TTU) Positive Series: Integral Test & p-series 3 March 204 8 / 8