By Chandan Ghughtyal
Srinivas Ramanujan was born on 22 December 1887 in Erode, Tamilnadu. His birthday is celebrated as the National Mathematics Day since 2012. This day is dedicated to honouring the life and legacy of Srinivasa Ramanujan. A self-taught genius and a true lover of numbers, Ramanujan gifted the world nearly 3,900 results, theorems and identities, many of which are still to prove for mathematic enthusiasts. It is rightly said ‘coming events cast their shadows before’. Ramanujan since his primary school level used to ask thought provoking questions like, how far are clouds from the earth, what is the quotient of zero divided by zero? He was an extraordinary person and used to practice only mathematics and patterns as a result he was only able to pass in mathematics in his senior secondary exam. But he was not disappointed and solved many unsolved questions of Prof HD Hardy, who was impressed with the mathematical aptitude of Srinivas Ramanujan. As a result, Prof Hardy invited Ramanujan to Cambridge and he did research with Prof Hardy. Due to his research work he was made a Fellow of the Royal Society, and was elected as the first Indian to become a Fellow of Trinity College, Cambridge.
Ramanujan gave the theta function which is used to study the growth of cancer cells. His pioneering work on partition formulas, π (pi) series, and infinite series laid a strong foundation for modern number theory. Today, these ideas continue to play a vital role in advanced fields such as cryptography, computer science, and theoretical physics, highlighting the enduring relevance and brilliance of Ramanujan.
On this auspicious occasion of the birthday of great son of Mother India, I would like to pay homage to his soul by explaining his incredible work, which is directly related to my classroom memory that left a lasting impact on my understanding of mathematics.
When I was a student of BSc at Kumaun University Campus, Almora, our professor once posed a seemingly simple question in class. He asked, “What is the sum of 10 consecutive natural numbers?”
We confidently replied, 55.
Then he asked for the sum of the first 100 natural numbers.
We answered, 5050 as we knew the formula for the sum of first n natural number is n (n+1)/2.
For the first 1000 natural numbers, after a short calculation, we said, 500,500.
Gradually, the numbers grew larger.
Then he asked, “What is the sum of the first trillion natural numbers?”
Some of us jokingly replied, “It must be something like a googolplex.”
Then came the question that silenced the room: “What if there are infinitely many natural numbers?”
We all answered instinctively, “The sum must be infinite.”
Professor smiled and calmly said: “The sum is minus one by twelve.”
And he wrote on the black board: 1+2+3+4+5+6+7+8+……………= – 1/12.
We were stunned.
How could the sum of all natural numbers, that grow endlessly result in a negative fraction? That time there was no internet, no Google to instantly verify such claims. Our curiosity turned into disbelief. The professor explained the difference between convergent and divergent series, but my mind remained fixed on that mysterious value . Since then, I always wonder how it can be negative. Will we get marks if we will write the answer in the exam? But the love for numbers automatically starts playing with the series.
Today, the explanation of this infinite series is as follows in the simplest terms:
Let the series T = 1-1+1-1+1-1+1………………
If added till the even point, the sum is 0 and if added till the odd point the sum is 1. It shows the sum oscillates between 0 and 1. The series is divergent as there is no fixed sum possible of the series. Using the partial sum property of divergent series the sum can be assigned a sum as the average of the two answers that is ½.
Now let’s assume the Infinite series U = 1-2+3-4+5-6+7………………… adding U with U by shifting the series one place to the right the sum can be written as follows:
2 U = 1+(-2+1)+(3-2)+(-4+3)_(5-4)+………….
2U = 1-1+1-1+1-1+1-1+………….
Now by partial sum the series the right-hand side is equal to ½.
Therefore, U=1/4
Now let us take the original series S=1+2+3+4+5+6+…………..
S-U= 1+2+3+4+5+6+…………..-(1-2+3-4+5-6+7…………)
S-U= 4+8+12+16+……………..
S-U= 4(1+2+3+4+5+………)
S-U = 4S
3S=-U
3S=-1/4
Therefore S = -1/12.
Today, the explanation of this infinite series can be presented in the simplest terms as follows:
Let the series
T = 1 – 1 + 1 – 1 + 1 – 1 + …
If the series is added up to an even number of terms, the sum is 0, and if it is added up to an odd number of terms, the sum is 1. This shows that the sum oscillates between 0 and 1. Since the series does not approach a fixed value, it is divergent. Using an averaging method based on partial sums (Cesàro summation), this series can be assigned the value 1/2.
Now, let us consider the infinite series
U = 1 – 2 + 3 – 4 + 5 – 6 + …
Adding the series U to itself after shifting it one place to the right, we obtain:
2U = 1 – 1 + 1 – 1 + 1 – 1 + …
As shown earlier, the series on the right-hand side is assigned the value 1/2.
Therefore, U = 1/4.
Now let us take the original series:
S = 1 + 2 + 3 + 4 + 5 + 6 + …
Subtracting U from S:
S – U = 4 + 8 + 12 + 16 + …
S – U = 4(1 + 2 + 3 + 4 + …) = 4S
Thus,
S – U = 4S
3S = -U
Since U = 1/4,
S = -1/12.
His work is dedicated to all those number lovers who always play with numbers like great Ramanujan.
Once, a student asked me whether I knew Ramanujan’s nested radical problem. I expressed my ignorance and later learnt it. Today it is an appropriate day to express my deepest gratitude to the great mathematician by presenting another fascinating result of Ramanujan.
Ramanujan’s nested radical problem:
Prove:
At first glance, this radical expression appears extremely complicated, and one naturally wonders how it can possibly lead to the value 3. Observing the right-hand side, we note that it is the square root of 9. The number 9 can be written as 1 + 8, where 8 is the product of 2 and 4. Here, 4 itself is the square root of 16. Continuing this pattern, 16 can be expressed as 1 + 15, and 15 is the product of 3 and 5, with 5 being the square root of 25. Through this continuous decomposition of numbers within the nested square roots, it becomes clear that the entire radical expression ultimately evaluates to 3.
Understanding the work of Srinivas Ramanujan, and reflecting on his work, is the true tribute to this great son of our country. On this auspicious day, I pay my humble homage to this extraordinary soul and wish each one of you a Happy National Mathematics Day. May we all learn to see the world through mathematical eyes.
(Chandan Ghughtyal is H.O.D. Mathematics, The Doon School, Dehradun.)
