Home » Sums » Page 2

Tag: Sums

Prove a formula for the sum of [(na)/b] where a,b are relatively prime integers

Let a,b be relatively prime positive integers (i.e., they have no common factors other than 1). Then we have the formula

    \[ \sum_{n=1}^{b-a} \left[ \frac{na}{b} \right] = \frac{(a-1)(b-1)}{2}. \]

The sum is 0 when b=1.

  1. Prove this result by a geometric argument.
  2. Prove this result by an analytic argument.

  1. Proof. We know by the previous exercise (1.11, #6) that

        \[ \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \begin{array}{l} \text{\# of lattice points inside a right triangle}\\ \text{of base } b \text{ and height } a \ (\text{letting } f(x) = \frac{ax}{b}).\end{array} \]

    Further, from this exercise (1.7, #4), we know

        \[ a(T) = I_T + \frac{1}{2} B_T - 1 \]

    where I_T = number of interior lattice points, and B_T = number of boundary lattice points. We also know by the formula for the area of a right triangle that

        \[ a(T) = \frac{1}{2} (ab). \]

    Thus, we have,

        \[ I = \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \frac{1}{2} (ab) - \frac{1}{2}B_T + 1. \]

    Then, to calculate B_T we note there are no boundary points on the hypotenuse of our right triangle (since a and b have no common factor). (This follows since if there were such a point then \frac{na}{b} \in \mathbb{Z} for some n < b, but then we would have a divides b, contradicting that they have no common factor.) Thus, B_T = a + b + 1. So,

        \begin{align*}  \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] &= \frac{1}{2} (ab) - \frac{1}{2} (a+b+1) +1 \\ &= \frac{1}{2} (ab - a - b + 1) \\ &= \frac{1}{2} (a-1)(b-1). \qquad \blacksquare \end{align*}

  2. Proof. To derive the result analytically, first, by counting in the other direction we have,

        \[ \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right]. \]

    Then,

        \begin{align*}  && \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right] &= \sum_{n=1}^{b-1} [a] - \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] - \sum_{n=1}^{b-1} 1 \\ \implies && 2 \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right] &= \sum_{n=1}^{b-1} [a] - \sum_{n=1}^{b-1} 1 = (b-1)a - (b-1)\\ \implies && \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] &= \frac{1}{2} (a-1)(b-1). \qquad \blacksquare \end{align*}

Evaluate if some formulas for finite sums are true or false

    Decide whether the following are true or false:

  1. \displaystyle{\sum_{n=0}^{100} n^4 = \sum_{n=1}^{100} n^4}.
  2. \displaystyle{\sum_{j=0}^{100} 2 = 200}.
  3. \displaystyle{\sum_{k=0}^{200} (2+k) = 2 + \sum_{k=0}^{100} k}.
  4. \displaystyle{\sum_{i=1}^{100} (i+1)^2 = \sum_{i=0}^{99} i^2}.
  5. \displaystyle{\sum_{k=1}^{100} k^3 = \left(\sum_{k=1}^{100} k \right) \left( \sum_{k=1}^{100} k^2 \right)}.
  6. \displaystyle{\sum_{k=0}^{100} k^3} = \left( \sum_{k=0}^{100} \right)^3}.

  1. True. Since 0^4 = 0 the extra term on the left has no impact on the sum. In other words,

        \[ \sum_{n=0}^{100} n^4 = 0^4 + \sum_{n=1}^{100} n^4 = \sum_{n=1}^{100} n^4. \]

  2. False. We know from here that \sum_{j=1}^{100} 1 = 100; hence,

        \[ \sum_{j=0}^{100} 2 = 2 \left(1 + \sum_{j=1}^{100} 1 \right) = 202. \]

    (The “problem” here is the extra term in the sum. The j=0 term actually contributes to the sum since 2 evaluated at j=0 is still 2.)

  3. False. This is just abusing the additive property of finite sums. In reality,

        \[ \sum_{k=0}^{200} (2+k) = \left( \sum_{k=0}^{200} 2 \right) + \left( \sum_{k=0}^{200} k \right) = 202 + \sum_{k=0}^{100} k. \]

  4. False. This is reindexing the sum incorrectly. Seeing, (i+1)^2 the idea is to replace a sum over i, by a sum over i+1 to get a simpler expression. (A quick and dirty way to see that this has not been reindexed correctly is by looking at the first few terms. The sum on the left starts 2^2 + 3^2 + 4^2 + \cdots, while the sum on the right starts 0^2 + 1^2 + 2^2 + \cdots. To do this correctly, we replace i+1 by i in the interior, and then have to add one to the top and bottom indices so that we have i+1=2 to 101 and can replace the i+1 by i. This is the way I think about it; there are probably better ways.) Anyway, reindexing properly we would get,

        \[ \sum_{i=1}^{100} (i+1)^2 = \sum_{i=2}^{101} i^2. \]

    Or we could actually expand out the (i+1)^2 squared term and use linearity to separate the sums and evaluate all of them separately, but that would be slower.

  5. False. Sums and powers just don’t work this way. (This is really claiming (1^3 + 2^3 + \cdots + 100^3) = (1+2 + \cdots +100)(1^2 + 2^2 + \cdots + 100^2).) From work we did earlier (see: sums of integers, sums of squares, and sums of cubes) we have,

        \[ \sum_{k=1}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=1}^n k \right) \left( \sum_{k=1}^n k^2 \right) = \left( \frac{n^2}{2} + \frac{n}{2} \right) \left( \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right). \]

    One can then plug in n=100 to see that we get 25502500 on the left, but 1708667500 on the right.

  6. False. Again, this just wrong. The claim is that (1^3 + 2^3 + \cdots + 100^3) = (1 + 2 + \cdots + 100)^3, which is of course false. More concretely, we know

        \[ \sum_{k=0}^{n} k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \]

    while

        \[ \left( \sum_{k=0}^n k \right)^3 = \left( \frac{n^2}{2} + \frac{n}{2} \right)^3.\]

    So, evaluating at n=100, we have 25502500 on the right and 128787625000 on the right.

Formula for sum of the reciprocals of integers

  1. Give a definition of \sum_{k=m}^{m+n} a_k.
  2. Prove the following is true for n \geq 1:

        \[ \sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}. \]


  1. We define

        \[ \sum_{k=m}^{m+n} a_k = a_m + a_{m+1} + \cdots + a_{m+n}.\]

  2. Proof. The proof is by induction. For the case n = 1 we have, on the left,

        \[ \sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{k=2}^2 \frac{1}{k} = \frac{1}{2}. \]

    On the right,

        \[ \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m} = \sum_{m=1}^2 \frac{(-1)^{m+1}}{m} = 1 - \frac{1}{2} = \frac{1}{2}. \]

    Thus, the formula holds for the case n = 1. Assume then that it holds for some n = j \in \mathbb{Z}_{\geq 1}. Then we have,

        \begin{align*}  &&\sum_{k=j+1}^{2j} \frac{1}{k} &= \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \\ \implies && \left(\sum_{k=j+1}^{2j} \frac{1}{k}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} &=  \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+1}^{2(j+1)} \frac{1}{k} &= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \frac{1}{j+1} + \left(\sum_{k=j+2}^{2(j+1)} \frac{1}{k} \right)&= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} + \frac{1}{2(j+1)} - \frac{1}{j+1} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} - \frac{1}{2(j+1)} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \sum_{m=1}^{2(j+1)} \frac{(-1)^{m+1}}{m}. \end{align*}

    Thus, the formula holds for j+1 if it holds for j; hence, we have shown that it holds for all n \in \mathbb{Z}_{\geq 1}. \qquad \blacksquare

Prove by induction a property of the alternating sum of odd integers

Prove that

    \[ \sum_{k=1}^{2n} (-1)^k (2k+1) = 2n. \]

This implies the sum is proportional to n with constant of proportionality 2.


Proof. The proof is by induction. For the case n = 1 we have, on the left,

    \[ \sum_{k=1}^{2n} (-1)^k (2k+1) = \sum_{k=1}^2 (-1)^k (2k+1) = -3 + 5 = 2. \]

On the right we have 2n = 2. Hence, the formula holds for this case.
Assume then that the formula holds for some n = m \in \mathbb{Z}_{>0}. Then,

    \begin{alignat*}{2}  \sum_{k=1}^{2m} (-1)^k (2k+1) = 2m \qquad &\implies & \sum_{k=0}^{2(m+1)} (-1)^k (2k+1) &= 2m - (2(2m+1)+1) + (2(2m+2)+1) \\ &&&= 2m - 4m - 3 + 4m + 5 \\ &&&= 2(m+1). \end{alignat*}\end{align*}

Thus, if the statement is true for m then it is true for m+1. Hence, we have established the statement is true for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare