擅长:python、mysql、java
<p>如果条件降得快到零</p>
<pre><code>def infinisum(f):
n, res = 0, f(0)
while True:
term = sum( f(k) for k in range(2**n,2**(n+1)) )
if (res+term)-res == 0:
break;
n,res = n+1, res+term
return res
</code></pre>
<p>那么你的特别系列是</p>
<pre><code>2*infinisum(lambda x: exp(-x**2)) - 1
</code></pre>
<p>分三步给出<code>1.772637204826652</code>,
或者一般来说,忘记了这个函数是对称的</p>
<pre><code>f = lambda x:exp(-x**2)
infinisum(f)+infinisum(lambda x : f(-1-x))
</code></pre>