Exercises on series

Test the following series

1. If a chessboard were to have wheat placed upon each square such that one grain were placed on the first square, two on the second, four on the third, and so on (doubling the number of grains on each subsequent square), how many grains of wheat would be on the chessboard at the finish?

2. $$\sum _{n=0}^{\mathrm{\infty }}\frac{n+5}{n^3-n^2+5}$$
3. $$\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}$$
4. $$\sum _{n=1}^{\mathrm{\infty }}[{(-\frac{1}{5})}^n+\frac{5}{\sqrt{n}}]$$
5. $$\sum _{n=1}^{\mathrm{\infty }}\frac{n+e^{-}n}{3n^2+5}$$
6. $$\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n-\sqrt{n}}{3n^4-1}$$
7. $$\sum _{n=1}^{\mathrm{\infty }}{n}^{100}{e}^{-n}$$
8. $$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{1/n}}{n!}$$
9. $$\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n}{n^3}$$
10. $$\sum _{n=1}^{\mathrm{\infty }}\frac{n!}{n^n}$$
11. $$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{\sqrt{n}}$$

Solutions

1. The number of grains of wheat on the square,

   1, 2, 4, 8, 16, 32, 64, 128, 256 ...

   form a geometric sequence, which grows at an increasing rate. Its sum can be obtained as

   $$s_n=\sum _{k=0}^{n-1}a{q}^k=a\frac{1-q^n}{1-q}$$

   where a is the first term of the series, q is the common ratio and n is the number of terms. Doing the calculations we have then 2⁶⁴ − 1. This  ■

2. Note that the general term a_n ~ 1/n², hence the series diverges. ■

3. ln n/n > 1/n then it diverges. ■

4. We study separately the behaviour of each series:

   Σ (−1/5)ⁿ

   is a geometric series with common ration −1/5, hence it converges. The second series is a generalized harmonic series with α = 1/2, hence it converges.  ■

5. The series is asymptotic to the harmonic series, hence it diverges.

   (n + e⁻ⁿ)/(3n² + 5) ~ 1/3 ⋅ 1/n ■

6. The series converves being asymptotic to the series Σ_n=1 1/n³ ⋅ 1/√n. Indeed we have

   (ln n −√n)/(3n⁴ − 1) ~ √n/3n⁴ = −1/3 ⋅ 1/(n³ √n) ■

7. It converges. We can use the ratio test whihc yields

   a_n+1/a_n = (n + 1)¹⁰⁰/e^{n + 1} = eⁿ/n¹⁰⁰ = 1/e ⋅ [(n + 1)/n]¹⁰⁰ ⟶ 1/e < 1 ■

8. It is an alternating series. We use the criterion of absolute convergence and study Σ |cos n|/n³ and since |cos n| ≤ 1 we have that the series is a minorant of Σ 1/n³ which we known to converge. Thus the series converges.  ■

9. We apply the ratio test

   $$\underset{n\to +\mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}=\underset{n\to +\mathrm{\infty }}{lim}\frac{\frac{(n+1)!}{(n+1{)}^{n+1}}}{\frac{n!}{n^n}}=\underset{n\to +\mathrm{\infty }}{lim}\frac{(n+1)!}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!}$$

   We have that (n + 1)^{n + 1} = (n + 1) ⋅ (n + 1)ⁿ and by the definition of factorial (n + 1)! = n! ⋅ (n + 1), then the limit is equal to

   $$\underset{n\to +\mathrm{\infty }}{lim}\frac{n!\cdot (n+1)}{(n+1{)}^n\cdot (n+1)}\cdot \frac{n^n}{n!}=\underset{n\to +\mathrm{\infty }}{lim}\frac{n^n}{(n+1{)}^n}=\underset{n\to +\mathrm{\infty }}{lim}{\left(\frac{n}{n+1}\right)}^n=\frac{1}{e}<1$$

   thus the series converges.  ■

10. Since n^1/n < n, we compare with the series Σ n/n!, which converges. ■

11. We use the Leibniz Criterion. We've that

    $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n}}=0$$

    and

    $$\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}$$

    hence the series converges.  ■

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