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i am solving some reccurence relations and I am getting lost and not sure where to go I have $$T(n)=T(n-1)+ 1/7^{(n-1)}\quad \text{ where } T(1)=1$$ Which I tried solving with something like this

T(1)=1
T(2)=T(1) + 1/7^(1)
T(3)=T(2) + 1/7^(2)
..
T(k)=T(n) + 1/7^(k)

I probably didnt completely understand how to solve it or I did something wrong. Couldnt find similar cases like these on the internet except for few where it is not well explained.

Please help me understand how to solve this case asymptotically and accurately.

Thank you

Matti P.
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ThePejnMan
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  • Your last equation is meaningless. –  May 26 '20 at 12:22
  • It's useful to write the sequence as $$ T_1 = 1 $$ $$ T_2 = 1 + \frac{1}{7} $$ $$ T_3 = 1 + \frac{1}{7} + \frac{1}{7^2} $$ Can you see the pattern? It's a sum ... – Matti P. May 26 '20 at 12:23

1 Answers1

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I think you mean ' in general case' by 'asymptotically'. Actually 'asymptotically' means at the infinity.

You can write your problem as

$$ T_n -T_{n-1}=\dfrac{1}{7^{n-1}} $$

And using what we call telescopic sum to find explicit expression of $T_n$.

$$T_N=\sum_{n=1}^N T_n-T_{n-1}=\sum_{n=1}^N\dfrac{1}{7^{n-1}}=...$$

Where the last sum is a geometric sum.

Can you finish the calculus ?

EDX
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