How to find 50-th term of this sequence and its sum

Question

Can someone help me with this excercise?

I recognize that the numerator is simply increasing by factors of 2, and every time a new number is written it begins from 1 again. I know that the denominator is similar in the sense that you write a natural number to the second power n times (where n=that natural number). I also know how to get the denominator of a50 (it is 100) and how to get the numerator (it is 9). I need help with (2) and (3). The problem is done without calculator.

Let $a_n$ be the $n$th term of the following sequence: $$\frac11,\frac14,\frac34,\frac19,\frac39,\frac59,\frac1{16},\frac3{16},\frac5{16},\frac7{16}\dots$$

1. Find $a_{50}$.

2. Find $\sum_{n=1}^{50}a_n$.

3. Find the maximal $n$ satisfying $a_n\ge\frac1{10}$.

Answers:

1. 0.09

2. 9.25

3. 190

Answer 1

To calculate the denominator, you need only do the following:

Figure out the denominator: $$50 = \dfrac{k(k-1)}{2}$$

$$k^2-k-100=0$$

You get $k \approx 10.5$. So, the denominator of $a_{50}$ is 100. Now, we want the numerator. $$\dfrac{10(10-1)}{2} = 45$$ so the 45th term is the last time you do not get a denominator of less than 100.

This means $a_{50} = \dfrac{2*(50-45)-1}{100} = \dfrac{9}{100}$.

Now, for the sum, you have:

$$\sum_{n=1}^{50} a_n = \sum_{n=1}^{45} a_n + \sum_{n=46}^{50} a_n = 1 + \left(\dfrac{1+3}{4}\right) + \left(\dfrac{1+3+5}{9}\right) + \cdots + \sum_{n=46}^{50} a_n$$

But we know that $k^2 = 1+3+5+\cdots + (2k-1)$. So, we wind up with:

$$\sum_{n=1}^{45} a_n = \underbrace{1+1+\cdots + 1}_{9\text{ times}} = 9$$

Then $$\sum_{n=1}^{50} a_n = 9+\dfrac{1+3+5+7+9}{100} = 9.25$$

Finally, to calculate the largest $n$ such that $a_n\ge \dfrac{1}{10}$, you need:

$$\dfrac{2k-1}{k^2} \ge \dfrac{1}{10}$$

Try for equality and go a little smaller.

$$k^2-20k+10=0$$

$$k \approx 19.5$$

So, we want: $$n\le \dfrac{19(20)}{2} = 190$$

Checking this out, we have:

$$n=190, a_{190} = \dfrac{37}{19^2} \approx 0.102$$

Answer 2

Hints:

For the second part, notice that $1 = \frac{1}{1}=\frac{1}{4}+\frac{3}{4}=\frac{1}{9}+\frac{3}{9}+\frac{5}{9}$. Make a prediction and attempt to prove it or find justification.

$1+3+5+\dots+(2k-1) = k^2$ Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$

For the third part, it is clear that the largest such $n$ would occur at the end of one of the "groups" of terms who share the same denominator. If it were the $k$'th such group, the last term would be of the form $\dfrac{2k-1}{k^2}$ and we ask what the largest $k$ would be still satisfying $\dfrac{2k-1}{k^2}\geq \dfrac{1}{10}$

Rearranging, we are asking what the largest $k$ is such that $20k-10\geq k^2$.

Once done, reinterpret the results in terms of $n$.

Answer 3

I think that the easiest and most straightforward way to solve this problem is to use a double sum. It's obvious that the denominators are squares, and the numerators are odd numbers. So we simply write the inner sum as $$\sum_{k=1}^m (2k-1),$$ and the outer sum is $$\sum_{m=1}^n \frac{1}{m^2}.$$ So the total sum is $$\sum_{m=1}^n \sum_{k=1}^m \frac{2k-1}{m^2}.$$ Then the $50^{\rm th}$ term corresponds to some choice of $m$ and $k$. How do we find it? Well, we know that the inner sum always has $m$ terms, so we want to choose the smallest $m$ such that $1 + 2 + \cdots + m > 50$, or equivalently, $m(m+1) > 100$. Clearly $m = 9$ is too small, but $m = 10$ works. So when $m = 9$ we have summed $9(10)/2 = 45$ terms, and for the next group with $m = 10$, we need $k = 5$ more to get to the $50^{\rm th}$ term. This means $$a_{50} = \frac{2(5)-1}{10^2} = \frac{9}{100}.$$

With the above in mind we can also compute $$\sum_{n=1}^{50} a_n = \sum_{m=1}^9 \sum_{k=1}^m \frac{2k-1}{m^2} + \sum_{k=1}^5 \frac{2k-1}{10^2}.$$ The first sum is simply $$\sum_{m=1}^9 \frac{1}{m^2} \sum_{k=1}^m (2k-1) = \sum_{m=1}^9 \frac{1}{m^2} \left( 2 \cdot \frac{m(m+1)}{2} - m \right) = \sum_{m=1}^9 1 = 9.$$ The second sum is $$\frac{1}{100} \sum_{k=1}^5 (2k-1) = \frac{1}{100} \left( 2 \frac{5(5+1)}{2} - 5 \right) = \frac{25}{100} = \frac{1}{4}.$$ Therefore, $$\sum_{n=1}^{50} a_n = \frac{37}{4}.$$

Finally, the largest $n$ such that $a_n \ge \frac{1}{10}$ is easily attained by noting that $2k-1$ is maximized when $k = m$, so all terms in the inner sum are smaller than $\frac{2m - 1}{m^2}$. So all terms will be less than $1/10$ when $$\frac{2m-1}{m^2} < \frac{1}{10}$$ or $$m > 10 + 3 \sqrt{10}$$ which implies $m \ge 20$. Consequently, when $m = 19$, the largest value of the inner sum is $$\frac{2(19)-1}{19^2} = \frac{37}{361} > \frac{1}{10}$$ and this is the final term exceeding $1/10$. This corresponds to $n = 19(19+1)/2 = 190$ (we did not add any additional $k$).

Answer 4

The terms with the same denominator $k^2$ come in groups of length $k$. Furthermore, the sum of the members of a group is always $1$.

Hence, the starting index of a group is a triangular number, $\dfrac{(k-1)k}2+1$ (for $1$-based indexing). The group containing $\color{green}{a_{50}}$ is with $k=10$, starting with $a_{46}$, i.e.

$$\frac{1}{100},\frac{3}{100},\frac{5}{100},\frac{7}{100},\color{green}{\frac{9}{100}},\cdots$$

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The sum up to this terms is $9$ for the $9$ first complete groups, plus $\dfrac{25}{100}$ for the partial $10^{th}$ group, hence

$$\sum_{n=1}^{50}t_n=\color{green}{\frac{37}4}.$$

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The last (largest) element of a group is

$$\frac{2k-1}{k^2}$$ and this exceeds $\dfrac1{10}$ until $k=19$. Hence, the last element of this group is

$$\color{green}{a_{190}}.$$