Summing Series: Solutions

Transcription

1 Sequences and Series Misha Lavrov Summing Series: Solutions Western PA ARML Practice December, 0 Switching the order of summation Prove useful identity 9 x x x x x x Riemann s zeta function ζ is defined to be the infinite sum ζ x x x x Find ζ ζ / / We define the n th harmonic number H n to be the value of the sum n, which has no closed form Express H in terms of H n H n n n H n n Concrete Mathematics Find again, in terms of H n H H Applying problem # from the next section, we get n n

2 Applying problem # from the next section yet again, the first sum in this result becomes n, so the whole thing simplifies to n Hn n 5 Prove useful identity by writing as can still save the day Let n n We have Hint: some things will go wrong, but you n n n n n Unfortunately, we don t now how to evaluate the very last sum here yet, but we do now how to evaluate the other two, and so we get n n n n n n Solving for n, we get the formula n nnn Find F, where F n is the n th Fibonacci number: F F and F n F n F n Hint: solve problem # in the next section first F F F F n F nf n F n F 7 ARML 978 Find the sum of the infinite series l Let l, so that we can relabel the inside sum as l l l l l This we can rewrite as l l

3 8 Putnam 00 Show that for each positive integer n, n! n lcm{,,, n/ } We show that both sides are equal by showing that for any prime p, p divides both sides an equal number of times Recall that the number of times p divides n! is n p n n We can rewrite this p p as n p and then reverse the summation to get :p <n/, :<n/p max{ : p < n/} But max{ : p < n/} is precisely the number of times p divides lcm{,,, n/ }, so the sum we ve ended up with is the number of times p divides the product on the left-hand side of the original equation Therefore we re done The method of differences Find the sum We can write as, which means that this sum simplifies to Find F, where F n is the n th Fibonacci number: F F and F n F n F n We can write F as F F, so this sum simplifies to F F F F F n F n F n F n F n F F n Wiipedia A well-nown but hard result is that this sum by using the upper bound π Find an approximation for /

4 and evaluating the sum on the right-hand side Bonus: what approximation for π do you get in this way? We have / // / / So this sum also telescopes to / 5/ 5/ 7/ 7/ 9/ / 5 This is actually very close to the truth: π 5 and 5 7 Relatedly, 0 is a pretty good approximation for π ARML 99 Let 5 99 x 99 Compute the integer x This is not a telescoping sum but a telescoping product We can write, so the product simplifies to and x , 5 a Prove useful identity 8 Pascal s identity states that r r r r, or r r r as r r So we have n r r n r b We have Use this, and useful identity 8, to derive useful identity We have which simplifies to nnn n n c Find a similar expression for, and use it with useful identity 8 to derive useful identity 7 Note: this method applies more generally to find the sum of any polynomial expression in The expression is, which can be found by repeatedly choosing the right binomial coefficient to subtract that will reduce the degree of the polynomial by From here, n n, n a Write the differences sinn sin n and cosn cos n in terms of sin n, cos n, and constants

5 We have: sinn sin n cosn cos n sin n cos cos n sin sin n cos sin n sin cos n cos n cos sin n sin cos n sin sin n cos cos n b Find a function fn such that fn fn sin n Begin by taing fn sin n sin cos n cos By the above identities, we ll get cos sin fn fn sin n cos n sin n cos n sin cos which simplifies to cos fn fn sin sin sin n sin n cos sin So we can adust our fn by multiplying it by sin, getting the new function fn sin n sin cos n cos In fact, fn further simplifies to cosn sin /, though we don t need that c Find a formula for sin Since sin f f, this sum telescopes to fn f, which simplifies to sin n n sin sin 7 VTRMC 0 Find Note that 8 Furthermore, the summand can be split into the partial fractions where Qx Q Q, x So the sum telescopes as x x Q Q Q Q5 Q Q Q Q, since as x, Qx 0 To find the value of the sum, all we have to do is evaluate Q Q 0 9 5

6 8 USAMO 99 For any set S, let σs and πs denote the sum and product, respectively, of the elements of S, with σ 0 and π Prove that σs πs n n n, n S [n] where the sum ranges over all subsets S of [n] {,,,, n} We can split the sum into two parts: sets S not including n, and sets S including n The first part is subsets of [n ], and sets in the second part can be written as T {n}, where T is a subset of [n ] So we have S [n] σs πs S [n ] S [n] σs πs T [n ] σt {n} πt {n} S [n ] We can simplify the second sum and rearrange to get σs πs σs n πs S [n ] T [n ] σs πs πt T [n ] σt n πt n The last summation can be solved directly: it is the product n, since when we expand this product, each term πt appears exactly once This product telescopes to exactly n, so we get the recurrence s n s n n, n where s n is the sum we are trying to evaluate From here, the statement we want can be shown by induction 9 a Find We have and in particular x x x x x x x x x x x x x, So this sum telescopes as all and in the end, only the first term remains b Show that odd

7 By similar logic, we have So each term in the right-hand sum corresponds to infinitely many terms of the lefthand sum: term is the sum of terms,,, 8, on the left-hand side, while term is the sum of terms,,,, on the left-hand side, term 5 is the sum of terms 5, 0, 0, 0, on the left-hand side, and so on Since each positive integer can be uniquely factored as an odd number times a power of, this means that each term of the left-hand sum is covered exactly once in this way, and the two sums are equal 7